Joelle can read 3 pages in 4 minutes, 4.5 pages in 6 minutes, and 6 pages in 8 minutes. What is the unit rate
Joelle can read 3 pages in 4 minutes, than means 3/4 = 0.75 pages per minute
4.5 pages in 6 minutes = 4.5/6 = 0.75 pages per minute
6 pages in 8 minutes = 6/8 = 0.75 pages per minute
The rate is contant and is equal to 0.75 pages per minute
Answer:
0.75 pages per minute
Question 12 points)The coordinates of Zare(2.1). What are the new coordinates of Zafter a dilationwith a scale factor of 2(4,2) (2,2) (2,4) (8,4) (1,0.5)
The result of Dilation is either enlargement or reduction of the original object or figure. This depend on the scale factor. A scale factor lesser than 1 results in reduction while a scale factor greater than 1 results in enlargement.
From the given information, the scale factor is 2 and the result would be an enlargement.
The new coordinates would be
(2 * 2, 1 * 2) = (4, 2)
Find the greatest common factor of the following monomialsgh^4 2g^3h
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.
Solve: (3x−2)^2 − 1 = 15. Enter the exact answers.
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]The table shows the amount of time four students practiced the trumpet one day.  Which list shows the names of the students in order from the least amount of practice time to the greatest amount of practice time? Gus, Ryan, Cole, Jacob ⊝ Cole, Jacob, Gus, Ryan ⊝ Ryan, Gus, Jacob, Cole ⊝ Ryan, Jacob, Cole, Gus
Express the fractional parts of the mixed numbers as decimal, and compare:
Cole : 2/3 = 0.66666
Gus : 1/2 = 0.5
Ryan: 1/4 = 0.25
Jacob : 7/12 = 0.58
From leat to greatest:
0.25 - 0. 5 - 0.58 - 0. 666
Ryan, Gus, Jacob, Cole
easiest stuff ever but I can't seem to get it
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.
what do you get for X on this proportion. 10/15 = 8/x
Can you please explain this as if I never knew this? I’m really struggling.
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%)
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%
Two similar cones have a scale factor of 8/27. What is the ratio of their volumes.
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.
The product of two numbers is 30. If one of the numbers is five over six, what is the other number?
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
Nina's grandmother deposits $3,000 into a savings account for her. The account pays 5.5% simple interest on an annual basis. If she does not add or withdraw money from the account, how much interest will she earn after 21 months? Round to the nearest cent
The amount of interest earned after 21 months is $288.75
Here, we want to calculate the amount of simple interest
To do this, we shall be using the simple interest formula
We have this as;
[tex]\begin{gathered} I\text{ = }\frac{P\times R\times T}{100} \\ \end{gathered}[/tex]Where;
I is the simple interest earned that we want to calculate
P is the principal which represents the amount deposited
R is the rate which is 5.5%
T is the time which is 21 months (in cases where time is not in years, we go on to divide the value of the month given by 12)
Thus, we have the substitution as follows;
[tex]I\text{ = }\frac{3000\times5.5\times21}{100\times12}\text{ = 288.75}[/tex]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.
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
Determine whether the following equation can be written as a linear function.x-9=7y3
the given equation ,
[tex]\frac{x-9}{3=7y}[/tex]The height power of the equation is 1.
therefore, the equation is linear.
Which number below is not an irrational number? A. π B. √2 C. √4 D. √6
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]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
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
00:00Express the repeating decimal 2.1as a fractiono2110o209199
Could I please with this math. I tried several times but still could not get all of them right
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.
How do you divide x by 6/18, and equate to 12/18.
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.
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)
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.
Which represents the following units in standard form? 2 hundred thousands, 1 ten thousand, 6 hundreds, 2 thousands. A- 2,162; B-200,162: C-210,610 D-212,600?
212600, represents the given units in the standard form.
from the right to left the first is ones, tens, hunders, thousand ,tenthousand and hundred thousand.
Which choice is equivalent to the product below when x>0√3/x*√x^2/12a. x/2b. √x/2c. x/4d. √x/2
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
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 :
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]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
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
How many square units will I need to paint the square pyramid, to the nearest squared unit?
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.
1/2 60 degrees, 30 degrees find x and y
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]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.
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]Solve for z, m and p. Type answers as whole numbers. For example, ifanswer is two type "2"
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]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
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
1 and 1/6divided by 1/12 in fraction form
The given expression is,
[tex]\frac{1\frac{1}{6}}{\frac{1}{12}}[/tex]On solving we have,
[tex]\begin{gathered} 1\frac{1}{6}=\frac{7}{6} \\ \frac{1\frac{1}{6}}{\frac{1}{12}}=\frac{7}{6}\times\frac{12}{1}=\frac{14}{1} \end{gathered}[/tex]