Polynomial Factoring
We are given the polynomial
[tex]P=d^2+12d+36[/tex]To factor the polynomial, we need to recall the identity:
[tex]\mleft(a+b\mright)^2=a^2+2ab+b^2[/tex]The trinomial (right side of the equation) consists in:
* the square of a variable
* twice the product of both variables
* the square of the other variable
The given expression has the corresponding terms:
* d^2 is the perfect square of d
* 36 is a perfect square, the square of 6
* 12d is twice 6d. Note the terms 6 and d are exactly the perfect squares, thus we can write:
[tex]P=d^2+12d+36=(d+6)^2[/tex]The third choice is correct
What is the lateral surface area of the swuare pyramid below?
Given:
There is a square pyramid given in the question.
Required:
We need to find the lateral surface area of given pyramid.
Explanation:
The Lateral area of pyramid is
[tex]A_L=2al[/tex]where a is the side of base which is 6 ft here
l is the slant height which is 15 ft here
now put all the values in formula
[tex]A_L=a*6*15=180\text{ ft}^2[/tex]Final answer:
the lateral surface area of given pyramid is 180 square ft
Place the steps for finding f^-1(x) in the correct order
We are given the following function:
[tex]f(x)=\sqrt[]{7x-21}[/tex]We are asked to determine its inverse function:
[tex]f^{-1}(x)[/tex]To do that, we will do the following change:
[tex]y=f(x)[/tex]We get:
[tex]y=\sqrt[]{7x-21}[/tex]Now, since we want to determine the inverse we will swap variables, like this:
[tex]x=\sqrt[]{7y-21}[/tex]Now we will solve for "y", first by squaring both sides:
[tex]x^2=7y-21[/tex]Now we will add 21 to both sides:
[tex]x^2+21=7y[/tex]Now we will divide both sides by 7:
[tex]\frac{1}{7}x^2+3=y[/tex]This is the inverse function, therefore, we can do the following change:
[tex]y=f^{-1}(x)[/tex]We get:
[tex]\frac{1}{7}x^2+3=f^{-1}(x)[/tex]Reduce the ratio to its lowest form. 63:9
Alex jogs 1 miles east, 5 miles north, and then 10 miles west.How far is Alex from his starting position, to the nearest tenth of a miles.
1 mils East + 5 miles North, and + 10 miles West.
It would be the same if I say the position of X = -9 and Y = 5
From Pythagorean theorem: a² + b² = c²
And in this case, c is the distance we want to calculate, a=X and b=Y
(-9)² + 5² = c²
81 + 25 = c²
106 = c²
c = √106 ≅ 10,3 miles
Which inequality is shown in this graph? (0, 3) (2,-3)
im not smart at all and i need my grade up in math or im screwed if i can just get a direct answer thatd be poggers. anyways: chris is mailing his friend a poster that has been rolled up in a long tube. he has a box that measures 20 inches by 8 inches by 4 inches. what is the maximum length the rolled poster can be?
the maximum length the rolled poster can be is 21.9089 inches
Christina and her 2 friends spent $25.02 at their favorite
restaurant. If they split the cost equally, about how much will they
each pay?
Answer:
Christina and her 2 friends will pay $8.34 each.
Step-by-step explanation:
Evaluate: can you please help me with these and explain
Given:
a)
[tex](3^{-3})(3^{-2})[/tex]b)
[tex](5^{-6})(5^{-3})[/tex]Required:
We need to eveluate both
Explanation:
a)
[tex]\frac{1}{3^3}*\frac{1}{3^2}=\frac{1}{243}[/tex]b)
[tex]\frac{1}{5^6}*\frac{1}{5^3}=\frac{1}{5^9}[/tex]Final answer:
[tex]\begin{gathered} \frac{1}{3^5}=\frac{1}{243} \\ \\ \frac{1}{5^9} \end{gathered}[/tex]Hello I’m having trouble getting the answer for this question.
We are asked to determine the surface area of the given figure. To do that we need to count the number of squares on each face, and we need to have into account that each square is equivalent to 1 square yards. Also, we need to have into account that each face repeats itself two times.
For the front face we have:
[tex]A_f=10yd^2[/tex]For the side face:
[tex]A_s=6yd^2[/tex]For the top face:
[tex]A_t=15yd^2[/tex]Now we add all the areas, multiplying each one by 2:
[tex]A=2A_f+2A_s+2A_t[/tex]Repalcing:
[tex]\begin{gathered} A=2(10)+2(6)+2(15)_{} \\ A=62yd^2 \end{gathered}[/tex]Therefore, the surface area of the figure is 62 square yards.
Use the given information to find the unknown value:y varies jointly as x, z, and w. When x = 2, z = 6, and w = 6. then y = 144. Find y when x = 6, z = 1, and w = 1.y =
Joint variation:
[tex]y=k\cdot x\cdot z\cdot w[/tex]Procedure:
0. We have to calculate the value of ,k ,with the first given values.
[tex]144=k\cdot2\cdot6\cdot6[/tex]2. Simplifying:
[tex]144=k\cdot2\cdot36[/tex][tex]144=k\cdot72[/tex][tex]k=\frac{144}{72}[/tex][tex]k=2[/tex]3. Finally, we find y using the second values given now that we now k.
[tex]y=2\cdot6\cdot1\cdot1[/tex]Answer: y = 12
Diana collected 5600 milliliters of rainwater on Saturday day. She collected 3.5 litter of rainwater on Sunday. How many total of milliliters did Daina collected on Saturday and Sunday?A. 910B. 4,060C. 4,600D. 9,100
Answer
D. 9,100
Explanation
What is given:
Diana collected 5600 milliliters of water on Saturday.
Diana collected 3.5 liters of water on Sunday.
What to find:
The total amount of water in milliliters Diana collected on Saturday and Sunday.
Step-by-step solution:
Step 1: Convert 3.5 liters of water collected on Sunday to milliliters.
Conversion factor: 1000 milliliters = 1 liter
Therefore, 3.5 liters = (3.5 liters/1 liter) x 1000 milliliters = 3500 milliliters
So 3500 milliliters of water were collected by Diana on Sunday
Step 2: Add the milliliters of water collected on Saturday and Sunday together.
Total milliliters collected = Milliliters of water collected on Saturday + Milliliters of water collected on Sunday
Total milliliters collected = 5600 milliliters + 3500 milliliters
Total milliliters collected = 9100 milliliters
Hence, the total milliliters of rainwater collected on Saturday and Sunday is 9100.
The correct answer is option D. 9,100
[tex] \frac{x - 1}{(x + 2)^{2} }[/tex]write the partial fraction decomposition.
Explanation
We are given the following expression:
[tex]\frac{x-1}{(x+2)^2}[/tex]We are required to determine the partial fraction decomposition of the given expression.
This is achieved thus:
We know that the partial fraction form of repeated roots is given as:
[tex]\frac{f(x)}{(x+a)^2}=\frac{A}{x+a}+\frac{B}{(x+a)^2}[/tex]Therefore, we have:
[tex]\frac{x-1}{(x+2)^2}=\frac{A}{x+2}+\frac{B}{(x+2)^2}[/tex]Next, we take the LCD and simplify as follows:
[tex]\begin{gathered} \frac{x-1}{(x+2)^{2}}=\frac{A}{x+2}+\frac{B}{(x+2)^{2}} \\ \frac{x-1}{(x+2)^{2}}=\frac{A(x+2)+B}{(x+2)^2} \\ \Rightarrow x-1=A(x+2)+B\text{ ----- \lparen equation 1\rparen} \end{gathered}[/tex]Next, we determine the values of A and B as follows:
[tex]\begin{gathered} x-1=A(x+2)+B \\ \text{ Let x = -2} \\ -2-1=A(-2+2)+B \\ -3=B \\ \therefore B=-3 \\ \\ From\text{ }x-1=A(x+2)+B \\ \text{ Let x = 0} \\ 0-1=A(0+2)+B \\ -1=2A+B \\ \text{ Substitute for B} \\ -1=2A-3 \\ 2A=2 \\ A=1 \end{gathered}[/tex]Therefore, the partial fraction becomes:
[tex]\begin{gathered} \frac{x-1}{(x+2)^2}=\frac{1}{x+2}+\frac{-3}{(x+2)^2} \\ \Rightarrow\frac{x-1}{(x+2)^2}=\frac{1}{x+2}-\frac{3}{(x+2)^2} \end{gathered}[/tex]Hence, the answer is:
[tex]\begin{equation*} \frac{1}{x+2}-\frac{3}{(x+2)^2} \end{equation*}[/tex]The question and the triangle are in the image.For part B I just forgot the formula I use to find the length of the segments. If you give me the formula that would be awesome so I can do it by myself. But part C I'll need help with
Solution:
Given:
Two transversals with four line segments.
[tex]AC,CE,BD,DF[/tex]Part A:
For the line segment AC, the length is the distance between points A and C.
[tex]\begin{gathered} A=(5,7) \\ C=(6,4) \\ \text{where;} \\ x_1=5,y_1=7 \\ x_2=6,y_2=4 \\ \\ \text{The distance betwe}en\text{ two points is given by;} \\ d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\ \text{Hence,} \\ d=\sqrt[]{(6-5)^2+(4-7)^2} \\ d=\sqrt[]{1^2+(-3)^2} \\ d=\sqrt[]{1+9} \\ d=\sqrt[]{10} \\ To\text{ the nearest hundredth,} \\ d_{AC}\approx3.16 \end{gathered}[/tex]For the line segment CE, the length is the distance between points C and E.
[tex]\begin{gathered} C=(6,4) \\ E=(7,1) \\ \text{where;} \\ x_1=6,y_1=4 \\ x_2=7,y_2=1 \\ \\ \text{The distance betwe}en\text{ two points is given by;} \\ d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\ \text{Hence,} \\ d=\sqrt[]{(7-6)^2+(1-4)^2} \\ d=\sqrt[]{1^2+(-3)^2} \\ d=\sqrt[]{1+9} \\ d=\sqrt[]{10} \\ To\text{ the nearest hundredth,} \\ d_{CE}\approx3.16 \end{gathered}[/tex]For the line segment BD, the length is the distance between points B and D.
[tex]\begin{gathered} B=(17,7) \\ D=(16,4) \\ \text{where;} \\ x_1=17,y_1=7 \\ x_2=16,y_2=4 \\ \\ \text{The distance betwe}en\text{ two points is given by;} \\ d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\ \text{Hence,} \\ d=\sqrt[]{(16-17)^2+(4-7)^2} \\ d=\sqrt[]{(-1)^2+(-3)^2} \\ d=\sqrt[]{1+9} \\ d=\sqrt[]{10} \\ To\text{ the nearest hundredth,} \\ d_{BD}\approx3.16 \end{gathered}[/tex]For the line segment DF, the length is the distance between points D and F.
[tex]\begin{gathered} D=(16,4) \\ F=(15,1) \\ \text{where;} \\ x_1=16,y_1=4 \\ x_2=15,y_2=1 \\ \\ \text{The distance betwe}en\text{ two points is given by;} \\ d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\ \text{Hence,} \\ d=\sqrt[]{(15-16)^2+(1-4)^2} \\ d=\sqrt[]{(-1)^2+(-3)^2} \\ d=\sqrt[]{1+9} \\ d_{}=\sqrt[]{10} \\ To\text{ the nearest hundredth,} \\ d_{DF}\approx3.16 \end{gathered}[/tex]Part B:
On the first transversal, the ratio of the lengths of the line segments formed on it is;
[tex]\begin{gathered} \frac{AC}{CE}=\frac{3.16}{3.16} \\ \\ \text{Hence, the ratio is;} \\ AC\colon CE=1\colon1 \end{gathered}[/tex]On the second transversal, the ratio of the lengths of the line segments formed on it is;
[tex]\begin{gathered} \frac{BD}{DF}=\frac{3.16}{3.16} \\ \\ \text{Hence, the ratio is;} \\ BD\colon DF=1\colon1 \end{gathered}[/tex]From the ratio of the lengths of each transversal, it is noticed that they are the same.
What is an equation of the line that passes through the point (6,3)(6,3) and is parallel to the line x+3y=24x+3y=24?
Hello there. To solve this question, we'll have to remember some properties about lines and properties of parallel lines.
We want to determine the equation of a line that passes through the point (6, 3) and is parallel to the line x + 3y = 24.
First, we'll rewrite the equation of this line, that was given in general form, into slope-intercept form:
For this, simply solve the equation for y
Subtract x on both sides of the equation
[tex]3y=24-x[/tex]Divide both sides of the equation by a factor of 3
[tex]y=-\dfrac{1}{3}x+8[/tex]We need this because in this form it is easier to find the slope of this line:
[tex]y=mx+b[/tex]So we find that
[tex]m=-\dfrac{1}{3}[/tex]Is the slope of this line.
A line that is parallel to this has the same slope, such that we can use the following equation to find the answer:
[tex]y=y_0+m(x-x_0)[/tex]Whereas (x0, y0) is the point the line passes through and m is the slope.
Plugging (x0, y0) = (6, 3) and m = -1/3 as we found, we get
[tex]y=3-\dfrac{1}{3}\cdot(x-6)[/tex]Multiply both sides of the equation by 3
[tex]\begin{gathered} 3y=9-(x-6) \\ \\ 3y=9-x+6 \\ \\ 3y=15-x \\ \end{gathered}[/tex]Add x on both sides of the equation
[tex]x+3y=15[/tex]This is the equation of the line passing through the desired point and is parallel to the line we had.
Select the property of equality that is used to generate each set of equations. Division Multiplication Addition Subtraction 4m = 60 4m + 2 = 60 = 2 7b + 9 = 22 7b +9-6 = 22-6 DO 9x - 4 = 33 9x - 4 + 3 = 33 + 3 у 3 у x 8 = 21 x 8 3
the given expression is,
4m = 60
4m ÷ 2 = 60 ÷ 2 ...........(Division)
7b + 9 = 22
7b + 9 - 6 = 22 - 6 ......... (subtraction)
9x - 4 = 33
9x - 4 + 3 = 33 + 3 ..............( Addition)
y/3 = 21
(y/3) x 8 = 21 x 8 ...............(multiplication)
Find (f/g)(x) when f(x)=3x+2 and g(x)=5x-6
We have this two functions:
[tex]\begin{gathered} f(x)=3x+2 \\ g(x)=5x-6 \end{gathered}[/tex]We have to calculate what f/g is.
[tex](\frac{f}{g})(x)=\frac{f(x)}{g(x)}=\frac{3x+2}{5x-6}[/tex]What is the domain of (f/g)(x)?
The domain is all the values of x for which the function has a defined value.
That means that if the denominator becames 0, the function is not defined.
This happens when 5x-6=0:
[tex]\begin{gathered} 5x-6=0 \\ 5x=6 \\ x=\frac{6}{5} \end{gathered}[/tex]Then, the domain is "all the real numbers different from 5/6".
are students desks a function of the classroom
Yes, because depending on the size of the classroom there may be more or less number of student's desks. In other words, the number of student's desks is a function of the number of m² in the classroom.
Solve the linear equation using equivalent equations to isolate the variable. Write your solution as aninteger, as a simplified fraction, or as a decimal number.13x + 5x++ 5 - -2AnswerKeypadKeyboard ShortcutsX=
The equation is given to be:
[tex]3x+5x+\frac{1}{6}=\frac{1}{2}[/tex]Step 1: Multiply each term by 6 to eliminate the fractions
[tex]\begin{gathered} 3x(6)+5x(6)+\frac{1}{6}(6)=\frac{1}{2}(6) \\ 18x+30x+1=3 \end{gathered}[/tex]Step 2: Add the common terms
[tex]48x+1=3[/tex]Step 3: Subtract 1 from both sides
[tex]\begin{gathered} 48x+1-1=3-1 \\ 48x=2 \end{gathered}[/tex]Step 4: Divide both sides of the equation by 48
[tex]\begin{gathered} \frac{48x}{48}=\frac{2}{48} \\ x=\frac{2}{48} \end{gathered}[/tex]Step 5: Reduce the fraction by dividing through by 2
[tex]\begin{gathered} x=\frac{2\div2}{48\div2} \\ x=\frac{1}{24} \end{gathered}[/tex]ANSWER
The solution is:
[tex]x=\frac{1}{24}[/tex]Which step contains and error and how would you correct that error? PLEASE HURRY FAST I WILL MARK BRAINIEST
1. Remove parenthesis: Distributive property (a(b+c)=ab+ac)
[tex]x-5x-5=3x+2[/tex]The error is in the line 1.2. Combine like terms:
[tex]-4x-5=3x+2[/tex]3. Substract 3x in both sides of the equation:
[tex]\begin{gathered} -4x-3x-5=3x-3x+2 \\ \\ -7x-5=2 \end{gathered}[/tex]4. Add 5 in both sides of the equation:
[tex]\begin{gathered} -7x-5+5=2+5 \\ \\ \\ -7x=7 \end{gathered}[/tex]5. Divide into -7 both sides of the equation:
[tex]\begin{gathered} \frac{-7}{-7}x=\frac{7}{-7} \\ \\ x=-1 \end{gathered}[/tex]
Find the value of x.
44 °
Х
x = [? ]°
Given the figure of a circle:
As shown, there is a cyclic quadrilateral
Every opposite angle are supplementary angles
So, the sum of the angles x and 44 is 180
So,
[tex]\begin{gathered} x+44=180 \\ x=180-44 \\ \\ x=136 \end{gathered}[/tex]So, the answer will be x = 136
(6,8) and (-12,-4) put in standard form
Problem
(6,8) and (-12,-4) put in standard form
Solution
The standard form for a line is given by:
y =mx +b
We can calculate the slope with this formula:
[tex]m=\frac{-4-8}{-12-6}=\frac{2}{3}[/tex]Then we can find the intercept using one point given for example x=6 and y=8 and we have:
8= 2/3*6 +b
b= 8- 4= 4
And our standard form would be given by:
y= 2/3 x +4
Consider function fand its graph.fte) =gsin(2)What is the amplitude of this function?
1) Considering that the general formula for a sinusoidal function is:
[tex]y=A\sin(Bx+C)+D[/tex]2) And that the amplitude is halfway between the highest and the lowest point of that curve.
3) We can tell that the amplitude is:
[tex]\frac{1}{2}[/tex]Find the surface area of each figure. Round answer to two decimal places. use 3.14 for pi
Solution
Step 1
Determine the number of shapes in figure 1
Figure 1 is made up of a cone and a hemisphere
Step 2
Write out the expression for the area of a cone and a hemisphere
[tex]\begin{gathered} \text{The area of a cone = }\pi\times r\times l \\ \text{The area of a hemisphere = 2}\times\pi\times r^2 \end{gathered}[/tex]Where
pi =3.14
r = 7 inches
l = 15 inches
Step 3
Substitute in the values and find the area of the shape
[tex]\begin{gathered} \text{Total area of the shape is given as} \\ \text{area of cone + area }of\text{ hemisphere} \\ =\text{ 2 x 3.14 }\times7^2+(3.14\text{ }\times7\times15) \\ =\text{ 307.72 + 329.7} \\ =637.42inch^2\text{ to 2 decimal places} \end{gathered}[/tex]Area = 637.42 square inches
Step 4
Determine the number of shapes in figure 2
Figure 2 is made up of a cone and a cylinder
Step 5
Write an expression for the area of a cylinder
[tex]\text{The area of a cylinder = 2}\times\pi\times r\times h\text{ + }\pi\times r^2[/tex]where h = 13yards
radius(r) = 11/2 = 5.5inches
l = ?
To find l, the slant height we use the Pythagoras theorem
so that
[tex]\begin{gathered} l^2=11^2+9^2 \\ l^2=202 \\ l\text{ =}\sqrt[]{202} \\ l\text{ = 14.2126704 in} \end{gathered}[/tex]Step 6
Substitute in the values and find the area of the shape
[tex]\begin{gathered} \text{Area of figure 2 is given as} \\ \pi\times r\times l\text{ + (2}\times\pi\times r\times h\text{ + }\pi\times r^2) \\ (3.14\text{ }\times\text{ 5.5}\times14.2126704)+((2\times3.14\times5.5\times13)+\text{ (3.14 }\times5.5^2) \end{gathered}[/tex][tex]\begin{gathered} 245.4528179+94.985+449.02=789.4578179Inches^2 \\ To\text{ 2 decimal places }\approx789.46inches^2 \end{gathered}[/tex]Area = 789.46 square inches
Choose the correct reasons for the given statements that are already there
Definition of a Parallelogram
1) Let's fill in the 2nd row of that table since we know that
Statement Reason
1. ABCD is a parallelogram Given
2. AB ≅ CE Given
3. CD ≅ AB Definition of a Parallelogram
Note that a parallelogram has 2 pairs of congruent and parallel line segments.
2) Thus, the answer is:
"Definition of a Parallelogram".
Use the table below to find each probability. The table gives information about students at one school.
Conditional Probabilities
The table gives us information about the favorite leisure activities of students at a school
We are required to find the probability of some events, given the occurrence of another event.
a) P(Sports | Female). It's the probability that a student likes sport if it's a female.
There are 334 female students, from which 39 like sports, thus:
[tex]P(S|F)=\frac{39}{334}=11.68\text{ \%}[/tex]b) P(Female | Sports). It's the probability that a student is known to like sports and it's also a female.
There are 106 students that like sports, from which 39 are female, thus:
[tex]P(F|S)=\frac{39}{106}=36.79\text{ \%}[/tex]c) P(Reading | Male). It's the probability that a male student likes reading.
There are 366 male students from which 76 like reading, thus:
[tex]P(R|M)=\frac{76}{366}=20.77\text{ \%}[/tex]d) P(Male | Reading). It's the probability that a student is known to like reading and it's also a male. There are 161 students who like reading out of which 76 are male, thus:also a male. d)
[tex]P(M|R)=\frac{76}{161}=47.20\text{ \%}[/tex]e) P(Hiking | Male). There are 366 male students out of which 58 like hiking, thus:H
[tex]P(H|M)=\frac{58}{366}=15.85\text{ \%}[/tex]f) P(Hiking | Female). There are 334 female students out of which 48 like hiking, thus:
[tex]P(H|F)=\frac{48}{334}=14.37\text{ \%}[/tex]g) P(Male | Shopping). There are 139 students who like shopping and from them, 68 are male, thus:39 students who like shopping and fr
[tex]P(M|S)=\frac{68}{139}=48.92\text{ \%}[/tex]h) P(Shopping | Female). There are 334 female students from which 71 like shopping, thus:
[tex]P(S|F)=\frac{71}{334}=21.26\text{ \%}[/tex]I) P(Phoning | Male). There are 366 male students from which 54 like phoning, thus:
[tex]P(P|M)=\frac{54}{366}=14.75\text{ \%}[/tex]ich 39 are female, thus:
S
Graph the line with the equation y = 4/3 x - 4
we are given the equation of a line, and we are asked to find the graph. To do that, let's remember that in order to find the graph of a line, we must find two points that lie on that line. The first number can be the y-intercept, let's remember the general equation for a line:
[tex]y=mx+b[/tex]where "m" represents the slope and "b" the intercept. In this case, the equation is:
[tex]y=\frac{4}{3}x-4[/tex]In this case, the intercept is b = -4. So our first point is (0,-4)
the second point can be found by making y = 0 and solving for "x", like this:
[tex]\begin{gathered} y=\frac{4}{3}x-4 \\ 0=\frac{4}{3}x-4 \end{gathered}[/tex]Adding 4 on both sides
[tex]4=\frac{4}{3}x[/tex]multiplying both sides by 3/4
[tex]\begin{gathered} 4(\frac{3}{4})=x \\ x=3 \end{gathered}[/tex]So the second point is (3,0). Now we locate the points and join them using a line, since the slope is a positive number (m=4/3) the line should go upwards. The graph is like this:
What is the magnitude of the vector a = (-3,5)?
In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data:
a = ( -3 , 5)
Step 02:
magnitude of the vector:
[tex]|a|\text{ = }\sqrt[]{x^2+y^2}=\sqrt[]{(-3)^2+5^2}=\sqrt[]{9+25}=\sqrt[]{34}[/tex]|a| = 5.83
The answer is:
|a| = 5.83
Question 81 pt 91 0 DetailsA die is rolled twice. What is the probability that a(n) 1 is rolled on the first roll and an even number on thesecond roll?The probability of rolling a(n) 1 on the first roll and an even number on the second roll is
Given:
A die is rolled twice.
To find:
What is the probability that 1 is rolled on the first roll and an even number on the second roll?
Explanation:
[tex]Probability\text{ of event= }\frac{favorable\text{ outcome}}{Total\text{ outcome}}[/tex]Solution:
A die is roll twice.
Each roll of the dice is an independent event. The probability of a 1 on one roll is 1/6 and the probability of an even number on the second roll is 3/6=1/2. Since, each roll is an independent event, the probability of a 1 on one roll and an even number on the second roll is
[tex]\frac{1}{6}\times\frac{1}{2}=\frac{1}{12}[/tex]Hence, these are the required probabilities.
For the triangle shown in the picture two angles are given find the third angle without using a protractor.
Let's use x for the missing angle. The sum of the internal angles of a triangle is equal to 180°. Then we have:
[tex]\begin{gathered} 180^{\circ}=x+29^{\circ}+109^{\circ}=x+138^{\circ} \\ 180^{\circ}=x+138^{\circ} \end{gathered}[/tex]Then if we substract 138° from both sides we get:
[tex]\begin{gathered} 180^{\circ}-138^{\circ}^{}=x+138^{\circ}-138^{\circ} \\ x=42^{\circ} \end{gathered}[/tex]Then the answer is 42°.
Find the annual rate of interest.Principal = 4600 rupeesPeriod5 yearsTotal amount =6440 rupeesAnnual rate of interest =%Stuck? Review related articles/videos or use a hint.
The compound interest is given, in general, by the next formula:
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]Where A is the amount, P is the initial amount, r is the rate of interest, n is the number of times the interest rate is applied in a period, and t is the numbers of periods.
Then, in our problem:
[tex]A=6440,P=4600,t=5,n=1[/tex]Solving the formula for r:
[tex]\begin{gathered} n=1 \\ \Rightarrow A=P(1+r)^t \\ \Rightarrow(1+r)^t=\frac{A}{P} \\ \Rightarrow r=-1+\sqrt[t]{\frac{A}{P}} \end{gathered}[/tex]Then, using the numerical values above:
[tex]r=-1+\sqrt[5]{\frac{6440}{4600}}=-1+\sqrt[5]{\frac{7}{5}}\approx-1+1.0696\approx0.07[/tex]Then, the rate of interest is approximately equal to 7%