(a) Find h(- 2), h(0), h(2) , and h(3) (b) Find the domain and range of h.(c) Find the values of x for which h(x) = 3 .(d) Find the values of x for which h(x) <= 3 .(e) Find the net change in h between x = - 3 and x = 3 .УА3h-303X
Answers
To find values of the function h at different x values, we go straight to that x value and find the corresponding y value of the graph.
From the graph, we have:
[tex]\begin{gathered} h(-2)=1 \\ h(0)=-1 \\ h(2)=3 \\ h(3)=4 \end{gathered}[/tex](b)The domain of a function is the set of x-values for which the graph of the function is defined.
The range of a function is the set of y-values for which the graph of the function is defined.
Looking at the graph, we see that from
x = - 3 to x = 4, the function is defined.
Also, from
y = - 1 to y = 4, the function is defined.
Thus, we can write the domain and range as >>>>>
[tex]\begin{gathered} D=-3\leq x\leq4 \\ R=-1\leq y\leq4 \end{gathered}[/tex](c)h(x) = 3 means y = 3
We will draw a horizontal line at y = 3 and see the points at which that line and curve crosses. Then, we will draw a perpendicular from that point to the x-axis. These are the values of x for which y = 3.
The graph:
We see from the graph drawn that for x = - 3, x = 2, and x = 4, the value of the function h is 3.
So,
[tex]x=-3,2,4[/tex](d)The values of x for which the function is ≤ 3 can be found by again drawing a line y = 3 and finding the places where the graph is BELOW that line.
Graph:
So, we can see that from x = - 3 to x = 2, the function is less than or equal to 3.
Thus,
[tex]-3\leq x\leq2[/tex](e)From x = - 3 to x = 3, the function changes several values. But the net change can be found by finding the respective values of the function at x = - 3 and at x = 3 and finding the difference.
At x = - 3, the function has a value of "3".
At x = 3, the function has a value of "4".
Thus, the net change is 4 - 3 = 1
Net Change = 1
Related Questions
what do you get for X on this proportion. 10/15 = 8/x
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don't understand letter Jcan u write the equation of the described function in standard form
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Answer:
y = x² - 5
Explanation:
We need to find a function such that when we replace each value of x, we get the respective value of y.
So, the function that describes the relationship is:
y = x² - 5
Because, when we replace x by 2, we get:
y = 2² - 5 = 4 - 5 = -1
If x = 3
y = 3² - 5 = 9 - 5 = 4
If x = 4
y = 4² - 5 = 16 - 5 = 11
If x = 5
y = 5² - 5 = 25 - 5 = 20
Two similar cones have a scale factor of 8/27. What is the ratio of their volumes.
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We have two cones.
They have a scale factor of 8:27.
This ratio relates elements in one dimension: radius, height, slant height.
This can be expressed for the height for example as:
[tex]\frac{h_2}{h_1}=\frac{8}{27}[/tex]If this is the ratio between the linear elements, then the ratio for the volume will be the cube of the linear ratio.
Then, we can find the ratio for the volumes as:
[tex]\frac{V_2}{V_1}=(\frac{8}{27})^3=\frac{512}{19683}[/tex]Answer: if the scale factor of the cones is 8/27, the ratio of their volumes is 512/19683.
Write the equation of the line with a slope of 3 that passes through the point (4,1).A. y = 3x + 13B. y= 3x - 11C. y= 3x - 1D. y=3x- 4
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The slope-intercept form of a line:
y = mx + b
where m is the slope and b is the y-intercept.
Replacing with m = 3 and point (4, 1) into the equation, we get:
1 = 3(4) + b
1 = 12 + b
1 - 12 = b
-11 = b
Then, the equation is:
y = 3x - 11
at a rental car company, a customer's daily rental charge D (in dollars) for m miles driven is determined by the function D=35+0.4m.PartA:List the values to complete the table.
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Answer:
39, 47, 55, 65, and
Explanation:
Given the below equation:
[tex]D=35+0.4m[/tex]When m = 10,
[tex]\begin{gathered} D=35+0.4(10) \\ =35+4 \\ \therefore D=39 \end{gathered}[/tex]When m = 30;
[tex]\begin{gathered} D=35+0.4(30) \\ =35+12 \\ D=47 \end{gathered}[/tex]When m = 50;
[tex]\begin{gathered} D=35+0.4(50) \\ =35+20 \\ D=55 \end{gathered}[/tex]When m = 75;
[tex]\begin{gathered} D=35+0.4(75) \\ =35+30 \\ D=65 \end{gathered}[/tex]When m = 100;
[tex]\begin{gathered} D=35+0.4(100) \\ =35+100 \\ D=75 \end{gathered}[/tex]How do you divide x by 6/18, and equate to 12/18.
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The Solution.
From the given question, we have
[tex]\frac{x}{(\frac{6}{18})}=\frac{12}{18}[/tex]The above equation is equivalent to
[tex]x\times(\frac{18}{6})=\frac{12}{18}[/tex][tex]\frac{18x}{6}=\frac{12}{18}[/tex][tex]3x=\frac{2}{3}[/tex]Cross multiplying, we get
[tex]3x\times3=2[/tex][tex]9x=2[/tex]Dividing both sides by 9, we get
[tex]\begin{gathered} \frac{9x}{9}=\frac{2}{9} \\ \\ x=\frac{2}{9} \end{gathered}[/tex]So, the value of x is 2/9.
Which choice is equivalent to the product below when x>0√3/x*√x^2/12a. x/2b. √x/2c. x/4d. √x/2
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The first step is to combine the terms by multiplication so that it becomes one term. We have
√(3/x) * √(x^2/12)
= √3x^2/12x
= √(x)/2
Option B is correct
easiest stuff ever but I can't seem to get it
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We are given that the function is related by the following
[tex]y=x+1[/tex]Let us find the missing values of x and y.
1st value:
Substitute y = 1 and solve for x
[tex]\begin{gathered} y=x+1_{} \\ 1=x+1_{} \\ x=1-1 \\ x=0 \end{gathered}[/tex]So, x = 0
2nd value:
Substitute x = -1 and solve for y.
[tex]\begin{gathered} y=x+1_{} \\ y=-1+1 \\ y=0 \end{gathered}[/tex]So, y = 0
3rd value:
Substitute x = 10 and solve for y.
[tex]\begin{gathered} y=x+1 \\ y=10+1 \\ y=11 \end{gathered}[/tex]So, y = 11
4th value:
Substitute y = 3 and solve for x.
[tex]\begin{gathered} y=x+1_{} \\ 3=x+1_{} \\ x=3-1 \\ x=2 \end{gathered}[/tex]Similarly, the rest of the values are calculated.
Greg needs to buy gas and oil for his cat. Gas costs $3.45 a gallon, and oil costs $2.41 a quart. He has $50 to spend. Write an equality where g is the number of gallons of gas he buys and q is the number of quarts of oil.
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3.45g + 2.41q = 50
Explanations:Price of one gallon of gas = $3.45
Price of one quart of oil = $2.41
Number of gallons of gas bought = g
Number of quarts of oil bought = q
Total amount spent = $50
Total amount of gas bought = (Price of one gallon of gas) x (Number of gallons of gas bought)
Total amount of gas bought = 3.45g
Total amount of oil bought = (Price of one quart of oil) x (Number of quarts of oil bought)
Total amount of oil bought = 2.41q
Total amount of gas bought + Total amount of oil bought = Total amount spent
3.45g + 2.41q = 50
Therefore, the required equation is:
3.45g + 2.41q = 50
Help I don’t understand and my last tutor didn’t even answer the problem
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Area and Definite Integrals
Definite integrals can be used to find the area under the graph of the functions and the x-axis. But integrals compute the areas above the x-axis as positive and below the x-axis as negative, so if we want to calculate the unsigned area, we must take care of the sign of the function in the given interval.
We are given the function:
[tex]f(x)=5-5x^2[/tex]And it's required to calculate the area between the x-axis and the function in the interval [0, 5]. We must find out if the graph crosses the x-axis to account for the change of signs.
To find possible roots of the function, we equate to zero:
[tex]\begin{gathered} 5-5x^2=0 \\ \text{Divide by -5:} \\ -1+x^2=0 \\ Add\text{ 1:} \\ x^2=1 \\ x=\sqrt[]{1}=\pm1 \end{gathered}[/tex]The root x = 1 belongs to the interval [0,5], thus the function changes signs at that point. An approximate graph of the function is shown below:
We can see the function is positive before x = 1 and negative after x = 1, so the integral for the last part should have the sign inverted as follows:
[tex]A=\int ^1_0(5-5x^2)dx-\int ^5_1(5-5x^2)dx[/tex]Calculate the indefinite integral:
[tex]\begin{gathered} I=\int ^{}_{}(5-5x^2)dx \\ I=5x-\frac{5}{3}x^3 \end{gathered}[/tex]Calculate the first integral evaluating it in 0 and 1:
[tex]\begin{gathered} (5\cdot1-\frac{5}{3}1^3)-(5\cdot0-\frac{5}{3}0^3) \\ 5-\frac{5}{3}=\frac{10}{3} \end{gathered}[/tex]Evaluate the second integral for x = 1 and x = 5:
[tex]\begin{gathered} (5\cdot5-\frac{5}{3}5^3)-\frac{10}{3} \\ (25-\frac{625}{3})-\frac{10}{3} \\ -\frac{560}{3} \end{gathered}[/tex]The required area is:
[tex]A=\frac{10}{3}+\frac{560}{3}=\frac{570}{3}=190[/tex]The area is 190 square units
Determine whether the functions are inversef(x) = 2(×-4)
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To determine the inverse function :
[tex]f(x)=\text{ 2(x-4)}[/tex][tex]\begin{gathered} f(x)=\text{ y} \\ y=2x-8 \\ y+8=2x \\ 2x=y+8 \end{gathered}[/tex]The inverse function is calculated by replacing the x with y variable
[tex]\begin{gathered} \frac{2x}{2}=\frac{y}{2}+\frac{8}{2} \\ x=\frac{y+8}{2} \\ f^{-1}(x)=\frac{x+8}{2} \end{gathered}[/tex]Therefore the inverse function :
[tex]f^{-1}(x)=\frac{x+8}{2}[/tex]
Could I please with this math. I tried several times but still could not get all of them right
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Recall that a median of a triangle is a line that goes from one vertex to the midpoint of the opposite side.
An angle bisector is a line that splits an angle into two equal angles.
An altitude of a triangle is a line that goes from one vertex to the opposite sides and the angle formed by the line and the side is a right angle.
Answer:
(a) Median of triangle FGH.
(b) Angle bisector of (c) Altutite of triangle ABC.
Which of the following is not a function?A. y-x=6B. y=2x²C. x= -2D. y+x=12
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Answer:
C. x= -2
Explanation:
Each of the options (A, B, and D) can be calculated for different values of x.
However, the equation below cannot:
[tex]x=-2[/tex]Therefore, x=-2 is not a function.
in a dog show there are 31 dogs competing in the terrier group the top three dogs will win a cash prize $500 and move on to compete for a place in the larger best in the show competition how many ways can the top three dogs be to determine if they're finishing position is not important
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The total number of dogs area 31.
Determine the number of ways for determining top 3 dogs inspite of their position.
[tex]\begin{gathered} ^{31}C_3=\frac{31!}{3!\cdot28!} \\ =\frac{31\cdot30\cdot29}{3\cdot2\cdot1} \\ =31\cdot5\cdot29 \\ =4495 \end{gathered}[/tex]So there are 4495 ways for selceting top 3 dogs from
Identify the vertices, foci and equations for the asymptotes of the hyperbola below. Type coordinates with parentheses and separated by a comma like this (x,y). If a value is a non-integer then type is a decimal rounded to the nearest hundredth. -4x^2+24x+16y^2-128y+156=0 The center is the point :
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ANSWER:
The center is:
[tex](3,4)[/tex]Vertex with larger y-value:
[tex](3,6)[/tex]Vertex with smaller y-value:
[tex](3,2)[/tex]Foci with larger y-value:
[tex](3,8)[/tex]Foci with smaller y-value:
[tex](3,0)[/tex]Equation of an asymptote:
[tex]y=0.5(x-3)+4[/tex]Where
[tex]\begin{gathered} a=0.5 \\ b=3 \\ c=4 \end{gathered}[/tex]EXPLANATION:
We have to take this equation into the general form of an hyperbola:
[tex]\frac{(x-h)^2}{b^2}-\frac{(y-k)^2}{a^2}=1[/tex]Where (h,k) is the center of the hyperbola.
We also know the vertices are:
[tex]\begin{gathered} (h,k+b) \\ (h,k-b) \end{gathered}[/tex]The foci are:
[tex]\begin{gathered} (h,k+2b) \\ (h,k-2b) \end{gathered}[/tex]And that the asymptotes are given by the expression:
[tex]y=\pm\frac{b}{a}(x-h)+k[/tex]Let's manipulate the equation:
[tex]\begin{gathered} -4x^2+24x+16y^2-128y+156=0 \\ \rightarrow16y^2-128y-4x^2+24x+156=0 \\ \rightarrow(16y^2-128y-4x^2+24x+156)\div4=0\div4 \\ \rightarrow4y^2-32y-x^2+6x+39=0 \\ \rightarrow4(y-4)^2-64-(x-3)^2+9+39=0 \\ \rightarrow4(y-4)^2-(x-3)^2-16=0 \\ \rightarrow4(y-4)^2-(x-3)^2=16 \\ \\ \rightarrow\frac{\mleft(y-4\mright)^{}_{}^2}{4}-\frac{(x-3)^2}{16}=1 \\ \\ \Rightarrow\frac{(y-4)^2_{}}{2^2}-\frac{(x-3)^2}{4^2}=1 \end{gathered}[/tex]From this general equation, we can conclude that the center is:
[tex](3.4)[/tex]Now, the vertices are:
[tex]\begin{gathered} (3,4+2)\rightarrow(3,6) \\ (3,4-2)\rightarrow(3,2) \end{gathered}[/tex]The foci are:
[tex]\begin{gathered} (3,4+4)\rightarrow(3,8) \\ (3,4-4)\rightarrow(3,0) \end{gathered}[/tex]And the equation of the asympotes are:
[tex]\begin{gathered} y=\pm\frac{2}{4}(x-3)+4 \\ \\ \rightarrow y=\pm\frac{1}{2}(x-3)+4 \end{gathered}[/tex]One of this asymptotes is:
[tex]y=\frac{1}{2}(x-3)+4[/tex]If 2 is added to a number and the sum is tripled, the result is 18 more than the number. Find the number.
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In this case, we are going to formulate an equation from the given statement. First we are told that 2 is added to a number that we are going to call x, then we can write:
x + 2
Then, we are told that this sum is tripled, that is we multiply the above expression by 3, then we get:
3(x + 2) = 3x + 6
And the above expression equals the number "x" plus 18, then we get:
3x + 6 = x + 18
From this expression, we can solve for x to get:
3x + 6 = x + 18
3x - x + 6 = x - x + 18
2x + 6 = 18
2x + 6 - 6 = 18 - 6
2x + 0 = 12
2x = 12
2x/2 = 12/2
x = 6
Then, the number is 6
Solve: (3x−2)^2 − 1 = 15. Enter the exact answers.
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x = 2 or x = -2/3
Explanation:[tex]\begin{gathered} \text{Given:} \\ (3x-2)^2\text{ - 1 = 15} \end{gathered}[/tex]To solve for x:
[tex]\begin{gathered} \text{add 1 to both sides:} \\ (3x-2)^2\text{ - 1 +1 = 15 + 1} \\ (3x-2)^2\text{ = 16} \\ \text{square root both sides:} \\ \sqrt[]{(3x-2)^2}\text{ = }\pm\sqrt[]{16} \end{gathered}[/tex][tex]\begin{gathered} 3x\text{ - 2 = }\pm4 \\ 3x\text{ = 2 }\pm4 \\ 3x\text{ = 2+4 }or\text{ }3x\text{ = 2 - 4} \\ 3x\text{ = 2 + 4} \\ 3x\text{ = 6} \\ x\text{ = 6/3} \\ x\text{ = 2} \end{gathered}[/tex][tex]\begin{gathered} 3x\text{ = 2 - 4} \\ 3x\text{ = -2} \\ x\text{ = -2/3} \\ \\ \text{Hence, x = 2 or x = -2/3} \end{gathered}[/tex]Solve for z, m and p. Type answers as whole numbers. For example, ifanswer is two type "2"
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Solution
Finding Z:
[tex]\begin{gathered} \text{ The sum of angles in a triangle is 180\degree} \\ \text{ Thus, we have:} \\ \\ 60\degree+90\degree+\angle Z\degree=180\degree \\ 150\degree+\angle Z=180\degree \\ \text{ Subtract 150 from both sides} \\ \\ \therefore\angle Z=180-150=30\degree \end{gathered}[/tex]Finding P:
[tex]\begin{gathered} \text{ Applying SOHCAHTOA, we have that:} \\ \tan60\degree=\frac{P}{\sqrt{3}} \\ \\ \therefore P=\sqrt{3}\times\tan60\degree \\ \\ P=\sqrt{3}\times\sqrt{3} \\ \\ P=3 \end{gathered}[/tex]Finding M:
[tex]\begin{gathered} \text{ Applying SOHCAHTOA once more, we have:} \\ \sin Z=\frac{\sqrt{3}}{m} \\ \\ \text{ We know Z= 30} \\ \\ \sin30\degree=\frac{\sqrt{3}}{m} \\ \\ \text{ We can rewrite this as:} \\ m=\frac{\sqrt{3}}{\sin30\degree} \\ \\ \text{ But,} \\ \sin30\degree=\frac{1}{2} \\ \\ \text{ Thus,} \\ m=\frac{\sqrt{3}}{\frac{1}{2}}=2\sqrt{3} \\ \\ m=2\sqrt{3} \end{gathered}[/tex]The product of two numbers is 30. If one of the numbers is five over six, what is the other number?
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The other number is 36
Explanation:Let the other number whose product with 5/6 gives 30 be x, then
(5/6)x = 30
Multiply both sides by 6/5
x = 30(6/5)
= 36
How many square units will I need to paint the square pyramid, to the nearest squared unit?
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Okay, here we have this:
Considering the provided figure and infomation, we are going to calculate the requested surface area, so we obtain the following:
So to calculate the requested area we are going to substitute in the following formula:
Total Surface Area= a(a + √(a^2 + 4h^2))
Replacing:
[tex]\begin{gathered} Total\text{ surface area}=8(8+\sqrt{8^2+4(4)^2}) \\ =8(8+\sqrt{64+4\cdot16}) \\ =8(8+\sqrt{64+64}) \\ =8(8+\sqrt{128}) \\ =8(8+8\sqrt{2}) \\ =64+64\sqrt{2} \\ \approx155units^2 \end{gathered}[/tex]Finally we obtain that you will need approximately 155 square units to paint the square pyramid.
00:00Express the repeating decimal 2.1as a fractiono2110o209199
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Determine whether the following equation can be written as a linear function.x-9=7y3
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the given equation ,
[tex]\frac{x-9}{3=7y}[/tex]The height power of the equation is 1.
therefore, the equation is linear.
The local department store is having their annual 25%-off sale.The regular prices for these purchases are: dress ($64.99), shoes ($46.00),coat ($149.98), and purse ($59.00). Find the total sale price for thesepurchases including a 5% sales tax for the final answer.
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the total sales price with regular prices = $64.99 + $46.00+$149.98+$59.00 = $319.97
after 25% - off sale, we have
total sales price = 319.97 - 25% of 319.97 = 319.97 - (0.25 x 319.97) = $239.98
after 5% sales tax
total sales price = $239.98 + 5% of $239.98 = $239.98+12 =251.98
Hence
the total sales price is $251.98
1/2 60 degrees, 30 degrees find x and y
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We will investigate the application of trignometric ratios.
There are three trigonometric ratios that are applied with respect to any angle in a right angle triangle as follows:
[tex]\begin{gathered} \sin \text{ ( }\theta\text{ ) = }\frac{P}{H} \\ \\ \cos \text{ ( }\theta\text{ ) = }\frac{B}{H} \\ \\ \tan \text{ ( }\theta\text{ ) = }\frac{P}{B} \end{gathered}[/tex]Where,
[tex]\begin{gathered} \theta\colon\text{ Any of the chosen angle of a right angle traingle except ( 90 degrees )} \\ P\colon\text{ Side opposite to the chosen angle} \\ B\colon\text{ Side adjacent/base to chosen angle} \\ H\colon\text{ Hypotenuse} \end{gathered}[/tex]We have two options to select our angle theta from:
[tex]\theta=\text{ 60 OR }\theta\text{ = 30}[/tex]We can choose either of the above angles. We will choose ( 30 degrees ); hence:
[tex]\begin{gathered} \theta\text{ = 30} \\ P\text{ = }\frac{1}{2}\text{ , B = y , H = x} \end{gathered}[/tex]We will use the trigonmetric ratios and evaluate each of the variables ( x and y ).
To determine ( x ) we can use the sine ratio as we have ( P ) and ( theta ) we can evaluate the hypotenuse as follows:
[tex]\begin{gathered} \sin (30)\text{ = }\frac{\frac{1}{2}}{x} \\ \\ x\text{ = }\frac{\frac{1}{2}}{\frac{1}{2}} \\ \\ x\text{ = 1}\ldots\text{Answer} \end{gathered}[/tex]To determine ( y ) we can use the tangent ratio as we have ( P ) and ( theta ) we can evaluate the Adjacent/base side as follows:
[tex]\begin{gathered} \tan (30)\text{ = }\frac{\frac{1}{2}}{y} \\ \\ y\text{ = }\frac{\frac{1}{2}}{\frac{\sqrt[]{3}}{3}} \\ \\ y\text{ = }\frac{1}{2\cdot\sqrt[]{3}}\ldots\text{Answer} \end{gathered}[/tex]Find the greatest common factor of the following monomialsgh^4 2g^3h
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Notice that:
[tex]\begin{gathered} gh^4=gh(h^3)^{} \\ 2g^3h=2(gh)(g^2) \end{gathered}[/tex]Therefore,
[tex]\text{GCF(gh}^4,2g^3h)=gh[/tex]Answer: gh.
Which number below is not an irrational number? A. π B. √2 C. √4 D. √6
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Irrational numbers usually possess an unending decimal. Irrational numbers cannot be expressed as a fraction for example x/y where y is not equal to 0. Example of irrational numbers are
[tex]\begin{gathered} \sqrt[]{2} \\ \pi \end{gathered}[/tex]The only number that is not an irrational number from the option is
[tex]\sqrt[]{4}[/tex]Given that tan A= 5/12 and tan B= -4/3 such that A is an acute angle and B is an obtuse angle find the value of,a) sin (A-B)b) cos (A+B)
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By trigonometric identity, solve for sin (A-B)
[tex]\begin{gathered} \sin (A-B)=\sin A\cos B-\sin B\cos A \\ \sin (A-B)=(\frac{5}{13})(-\frac{3}{5})-(\frac{4}{5})(\frac{5}{13}) \\ \sin (A-B)=-\frac{3}{13}-\frac{4}{13} \\ \sin (A-B)=-\frac{7}{13} \end{gathered}[/tex]Therefore, sin(A-B) = -7/13.
Solve for cos(A+B)
[tex]\begin{gathered} \cos (A+B)=\cos A\cos B-\sin A\sin B \\ \cos (A+B)=(\frac{12}{13})(-\frac{3}{5})-(\frac{5}{13})(\frac{4}{5}) \\ \cos (A+B)=-\frac{36}{35}-\frac{4}{13} \\ \cos (A+B)=-\frac{608}{455} \end{gathered}[/tex]Therefore, cos(A+B) = -608/455.
Find the difference in the volume and total area of a cylinder with both a radius and height of 1.r = 1, h = 1The number of sq units of the total area exceeds the number of cu. units in the volume by
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Answer:
The number of sq units of the total area exceeds the number of cubic units in the volume by 9.43
Explanation:
Given that the radius and the height of the cylinder is;
[tex]\begin{gathered} r=1 \\ h=1 \end{gathered}[/tex]Recall that the formula for the total surface area of a cylinder is;
[tex]A=2\pi r(h+r)[/tex]and the volume of a cylinder can be calculated using the formula;
[tex]V=\pi r^2h[/tex]Substituting the given values;
The surface area is;
[tex]\begin{gathered} A=2\pi r(h+r) \\ A=2\pi(1)(1+1) \\ A=2\pi(2) \\ A=4\pi \\ A=12.57\text{ sq units} \end{gathered}[/tex]The Volume is;
[tex]\begin{gathered} V=\pi r^2h \\ V=\pi(1)^2(1) \\ V=\pi \\ V=3.14\text{ cubic units} \end{gathered}[/tex]the difference between the volume and the total area of the cylinder is;
[tex]\begin{gathered} difference=A-V \\ d=12.57-3.14 \\ =9.43 \end{gathered}[/tex]Therefore, the number of sq units of the total area exceeds the number of cubic units in the volume by 9.43
Can you please explain this as if I never knew this? I’m really struggling.
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To determine the value of cos (A) using pythagoras theorem:
Using pythagoras theorem:
[tex]\begin{gathered} \text{Hypotenuse = 50} \\ \text{Opposite = 30} \\ \text{Adjacent = 40} \end{gathered}[/tex][tex]\begin{gathered} \text{Hyp}^2=\text{opp}^2+\text{adj}^2 \\ to\text{ CONFIRM } \\ 30^2+40^2=50^2 \\ 900+1600=2500 \end{gathered}[/tex][tex]\begin{gathered} Cos\text{ A = }\frac{adj}{\text{hyp}} \\ When\text{ A = angle subtends by the triangle} \\ adj=30 \\ hyp=50 \\ \text{Cos A=}\frac{30}{50} \\ \text{CosA=}\frac{3}{5} \end{gathered}[/tex]Therefore the value of CosA = 3/5
I currently have a 95.52 % in my math class and my final exam is today.I took a practice final as reference and got a 70%.what percentage do I have to get on the final, to have a passing grade (passing grade is at least 70%, I currentlyhave 95.52%)
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The current score is given as 95.52%.
The passing grade is required to be a 70%.
This means the score on the final that is needed to achieve a 70% passing grade would be calculated as follows;
[tex]\begin{gathered} \frac{95.52+x}{2}=70 \\ x=\text{the required score in the final exam} \\ \text{Cross multiply;} \\ 95.25+x=140 \\ x=140-95.52 \\ x=44.48 \end{gathered}[/tex]The answer means that, on the final exam you must score at least (nothing less than) 44.48% in order to have a passing grade of 70%