Square roots are rational only when has perfect square factors.
√10 = √(2*5) = √2 * √5
2 and 5 are not perfect squares, then √2 and √5 are irrational. In consequence, √10 is also irrational.
Write an equation for a function that gives thr value in the table. Use the equation to find the value of y when x= 12.
the First, we figure out the equation of the line.
[tex]\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}[/tex]We pick point (8,27) as point 1 and (10,35) as point 2.
Therefore equation of line =
[tex]\begin{gathered} \frac{y-27_{}}{x-8_{}}=\frac{35-27_{}}{10-8_{}} \\ \frac{y-27_{}}{x-8_{}}=\frac{8}{2} \end{gathered}[/tex]Cross multiplying, we have
8x -64 = 2y - 54
Adding 54 to both sides, we have:
8x - 10 = 2y
Dividing both sides by 2, we have:
y = 4x -5
Next, we substitute the value of x with 12 to get:
y = 4(12) - 5
y = 48 - 5
y = 43
Express the confidence interval 0.25 < p < 0.43 in the form of p ± E.
The confidence interval written as 0.25 < p < 0.43 in the form p ± E is; CI = 0.34 ± 0.09
How to express confidence Interval?The general formula for confidence interval is;
CI = x' ± z(s/√n)
where;
CI is confidence interval
x' is sample mean
z is confidence level value
s is sample standard deviation
n is sample size
We are given the confidence interval as;
0.25 < p < 0.43
To write it in the form p ± E means that;
- p = (up confed limit) + (low conf limit)/2
- E= (up confed limit) - (low confed limit)/2
p = (0.25 + 0.43)/2
p = 0.34
E = (0.43 - 0.25)/2
E = 0.09
Thus;
CI = 0.34 ± 0.09
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describe the domain of the function f(x;y)= ln(4-x-y)
Domain of the given function is x∈(-2,∞)
Step-by-step explanation:
The given function is y=\ln(x+2)y=ln(x+2)
Domain is the set of x values for which the function is defined.
And we know that logarithm function is defined only for values greater than zero.
Therefore, for domain we have
x + 2 >0
x > -2
Hence, the domain of the
The domain of the function
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What Is Domain and Range in a Function?
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Chapter 7 / Lesson 3
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What are the domain and range of a function? What are the domain and range of the graph of a function? In this lesson, learn the definition of domain and range as it applies to functions as well as how it applies to graphs of functions. Moreover, there will be several examples presented of domain and range and how to find them.
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Omor is preparing the soil in his garden for planting squash. The directions say to use 4 poundsof fertilizer for 160 square feet of soil The area of Omar's garden is 200 square feet.How much fertilizer is needed for a 200 square-foot garden?1
So the formula say 4 pounds of fertilizer for 160 square feet and he has to made for 200 square feet so we can made a rule of 3 so:
[tex]\begin{gathered} 4\to160 \\ x\to200 \end{gathered}[/tex]and the equation will be:
[tex]\begin{gathered} x=\frac{200\cdot4}{160} \\ x=5 \end{gathered}[/tex]So he need 5 pounds of fertilizer
Answer the following. (a) Find an angle between 0 and 2π that is coterminal with -13T/12 (b) Find an angle between 0° and 360° that is coterminal with 810°. Give exact values for your answers.
(a) Find an angle between 0 and 2π that is coterminal with -13π/12.
To get coterminal angles, we simply have to add or subtract 2π. In this problem, we are looking for a coterminal angle that is between 0 and 2π, so we will add 2π to -13π/12:
[tex]\frac{-13\pi}{12}+2\pi=\frac{11\pi}{12}[/tex](b) Find an angle between 0° and 360° that is coterminal with 810°.
Similarly to the previous exercise, we will substract 360° to 810° to find a coterminal angle that is between 0 and 360°:
[tex]\begin{gathered} 810\text{\degree}-360\text{\degree}=450\text{\degree} \\ 450\text{\degree}-360\text{\degree = 90\degree} \end{gathered}[/tex]ANSWER
a. 11π/12
b. 90°
solve each equation for y=. Without graphing, classify each system as having one solution, no solution, or infinitely many solutions.
x+y=3
y=2x-3
Answer:
The answer would be y=3, and there is only one solution.
Step-by-step explanation:
In the first expression, x+y=3, we can rearrange it to get it in terms of x so we can substitute it for x in the second expression.
x+y=3
Subtract y from both sides: x=-y+3
Substitute x=-y+3 into the second expression: y=2(-y+3)+3
Distribute the 2: y=-2y+6+3
Simplify the right side: y=-2y+9
Add 2y to both sides: 3y=9
Divide by 3: y=3
Since there is a single y coordinate, that means that there is only one solution.
The points (−5, -5) and (r, 1) lie on a line with slope 1/2. Find the missing coordinate r.
Solution:
The slope is expressed as
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ where \\ (x_1,y_1)\text{ and} \\ (x_2,y_2)\text{ are the coordinates of points through which the line passes} \end{gathered}[/tex]Given that the points (-5, -5) and (r, 1) lie on the line with slope 1/2, this implies that
[tex]\begin{gathered} x_1=-5 \\ y_1=-5 \\ x_2=r \\ y_2=1 \end{gathered}[/tex]By substituting these valus into the slope formula, we have
[tex]\begin{gathered} \frac{1}{2}=\frac{1-(-5)}{r-(-5)} \\ \Rightarrow\frac{1}{2}=\frac{1+5}{r+5} \\ cross-multiply, \\ r+5=2(1+5) \\ \Rightarrow r+5=12 \\ add\text{ -5 to both sides of the equation,} \\ r+5-5=12-5 \\ \Rightarrow r=7 \end{gathered}[/tex]Hence, the missing coordinate r is evaluated to be
[tex]7[/tex]For the following set of data, find the number of data within 1 population standarddeviation of the mean.68, 68, 70, 61, 67, 71, 63, 67
Given the following set of numbers,
[tex]68,\text{ 68, 70, 61, 67, 71, 63, 67}[/tex]Where the (n) number of data is 8, the mean is,
[tex]\bar{x}=\frac{x_1+x_2+..._{}+x_n}{n}=\frac{68+68+70+61+67+71+63+67}{8}=66.875[/tex]The standard deviation is 3.36
Hence, the interval that is 1 population within the mean is given by
[tex](66.88-3.36,66.88+3.36)=(63.52,70.24)[/tex]Of all the data only 71, 61, and 63 are not an element of the interval (63.52,70.24)
The total number of data is 8.
Hence, the total number of data within 1 standard deviation of the mean is 5
Four people buy raffle tickets and place all of their tickets into a hat. Raphael bought 8 tickets, Leonardo bought 2 tickets, Michelangelo bought 5 tickets, and Donatello bought 11 tickets. If a person randomly selects one raffle ticket from the hat, what is the probability that Raphael purchased the ticket? Write your answer as a decimal and round three decimal places.
0.308
Explanation
The probability of an event is the number of favorable outcomes divided by the total number of outcomes possible,so
[tex]P(A)=\frac{favorable\text{ outcomes }}{total\text{ outcomes}}[/tex]so
Step 1
favourable outcome is the result that is desired, so as the result is a that Raphael purchase the ticket
a) Let
[tex]favorable\text{ outcome=}8[/tex]and , the total outcome is the total ticket,so
[tex]\begin{gathered} total\text{ outcome=Raphael+Leonardo+Michelangelo+Donatello} \\ total\text{ outcome=8+2+5+11=26} \end{gathered}[/tex]b) now, replace in the formula:
[tex]\begin{gathered} P(A)=\frac{favorable\text{ outcomes }}{total\text{ outcomes}} \\ P(A)=\frac{8}{26} \\ P(A)=0.3076923 \\ rounded \\ P(A)=0.308 \end{gathered}[/tex]therefore,the answer is 0.308
I hope this helps you
Ivan took out a loan for 6700 that charges an annual rate of 9.5% compounded quarterly. Answer each part.
We will have the following:
a) The amount after one year will be:
[tex]\begin{gathered} A=6700(1+\frac{0.095}{4})^{4\ast1}\Rightarrow A=7359.53647... \\ \\ \Rightarrow A\approx7359.54 \end{gathered}[/tex]So, the amount after 1 year will be approximately $7359.54.
b) The effective annual interest rate will be:
[tex]eair=(1+\frac{0.095}{4})^4-1\Rightarrow eair=0.0984382791...[/tex]So, the effective annual interest rate will be approximately 9.84%.
Find the slope-intercept equation of the line that passes
through (-4, 2) and has a slope of 1/4
Start with y=mx+b and using the slope and the point, find b.
By knowing the slope, you know y = (1/4) x + b.
If you substitute in (-4,2), you'd have: 2 = (1/4)•(-4) + b
2 = -1 + b
3 = b
So your equation is y = 1/4 x + 3.
A triangle has vertices on a coordinate grid at P(4, -9), Q(-1, -9), and R(4,6).What is the length, in units, of PQ?
Given:
The coordinates of point P, (x1, y1)=(4, -9).
The coordinates of point Q, (x2, y2)=(-1, -9).
The length of PQ can be calculated as,
[tex]\begin{gathered} PQ=\sqrt[]{(x2-x1)^2+(y2-y1)^2} \\ PQ=\sqrt[]{(-1-4)^2+(-9-(-9))^2} \\ PQ=\sqrt[]{(-5)^2+0} \\ PQ=\sqrt[]{5^2} \\ PQ=5 \end{gathered}[/tex]Therefore, the length PQ
$3,700 for 2% for 4 yearswhat is the simple interest?what is the total amount?
Here, we want to get the amount on the simple interest
Mathematically, this is the sum of the amount deposited and the interest accurred
For the interest, we use the formula for simple interest as follows;
[tex]\begin{gathered} I\text{ = }\frac{PRT}{100} \\ \\ P\text{ is the amount deposited = \$3,700} \\ R\text{ is the rate which is 2\%} \\ T\text{ is time which is 4 years} \\ \text{Substituting these values;} \\ I\text{ = }\frac{3700\times2\times4}{100}\text{ = \$296} \end{gathered}[/tex]So, we simply add this to the principal to get the amount
[tex]\begin{gathered} \text{Amount = Principal + Interest} \\ =\text{ \$3,700 + \$296 = \$3,996} \end{gathered}[/tex]Round 6,752 to the nearest ten and nearest hundred.
Given the number:
6752
i) Round to the nearest ten:
To round to nearest ten means to rou
I need help I'm still having a bit of trouble with multiplying radicals
The area of the rectangle is;
[tex]36\sqrt[]{3\text{ }}\text{ sq cm}[/tex]Here, we want to find the area of the rectangle
To do this, we will have to multiply the length of the sides
Mathematically, we have this as follows;
[tex]\begin{gathered} \text{Area = Length }\times\text{ Width} \\ \text{Area = 6 cm }\times\text{ 6}\sqrt[]{3\text{ }}\text{ cm} \\ =\text{ 36}\sqrt[]{3}\text{ sq cm} \end{gathered}[/tex][tex]\begin{gathered} 6\times6\text{ = 36} \\ \sqrt[]{6}\text{ }\times\text{ 6 = 6}\sqrt[]{6} \\ \sqrt[]{6}\text{ }\times\text{ }\sqrt[]{6}\text{ = 6} \\ 6\sqrt[]{6}\text{ }\times\text{ 6}\sqrt[]{6}\text{ = (6}\times6)\times(\sqrt[]{6\text{ }}\text{ }\times\text{ }\sqrt[]{6}\text{ ) = 36}\times6\text{ = 216} \\ \sqrt[]{3}\text{ }\times\text{ }\sqrt[]{2\text{ }}\text{ = }\sqrt[]{\text{ (3}\times2)}\text{ = }\sqrt[]{6} \\ 7\sqrt[]{3}\text{ }\times\text{ 8}\sqrt[]{7\text{ }}\text{ = (7}\times8\text{ )}\times\text{ (}\sqrt[]{3}\text{ }\times\sqrt[]{7\text{ }}\text{ ) = 56}\sqrt[]{21} \end{gathered}[/tex]a. in radiansb.in degreesHint:HEARRERANote: You can earn partial credit on this problem.Preview My AnswersSubmit Answers
EXPLANATIONS:
Given;
We are given the following expression;
[tex]arctan(\frac{1}{\sqrt{3}})[/tex]Required;
We are required to find the angle measure of this in both radians, and degrees.
Step-by-step solution;
For the angle whose tangent is given as 1 over square root of 3, on the unit circle, we would have
[tex]\begin{gathered} tan\theta=\frac{1}{\sqrt{3}} \\ Rationalize: \\ \\ \frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ =\frac{\sqrt{3}}{\sqrt{3}\times\sqrt{3}} \\ \\ =\frac{\sqrt{3}}{3} \end{gathered}[/tex]On the unit circle, the general solution for this value as shown would be;
[tex]tan^{-1}(\frac{\sqrt{3}}{3})=\frac{\pi}{6}[/tex]To convert this to degree measure, we will use the following equation;
[tex]\frac{r}{\pi}=\frac{d}{180}[/tex]We now substitute for the value of r;
[tex]\begin{gathered} \frac{\frac{\pi}{6}}{\pi}=\frac{d}{180} \\ \\ \frac{\pi}{6}\div\frac{\pi}{1}=\frac{d}{180} \\ \\ \frac{\pi}{6}\times\frac{1}{\pi}=\frac{d}{180} \\ \\ \frac{1}{6}=\frac{d}{180} \end{gathered}[/tex]We now cross multiply;
[tex]\begin{gathered} \frac{180}{6}=d \\ \\ 30=d \end{gathered}[/tex]Therefore;
ANSWER:
[tex]\begin{gathered} radians=\frac{\pi}{6} \\ \\ degrees=30\degree \end{gathered}[/tex]I don't understand this, can you hell me solve this please?
We will investigate the angle measures and the properties involved with a pair of parallel lines.
We are given two pairs of parallel lines, namely:
[tex]\begin{gathered} L\text{ }\mleft\Vert\text{ m }\mright? \\ a\text{ }\mleft\Vert\text{ b}\mright? \end{gathered}[/tex]The angle properties that are used in consequence of parallel lines are of the following:
[tex]\text{Alternate Angles , Complementary Angles , Supplementary Angles, Corresponding Angles}[/tex]Each of the above property describes a relationship between two angle measures. That is how two angles are related to one another in consequence of the parallel lines.
The angle measures are classified into two types as follows:
[tex]\begin{gathered} \text{Interior Angles} \\ \text{Exterior Angles} \end{gathered}[/tex]5. The function w(x) = 70x represents the number of words w(x) you can type in x minutes. SHOW ALL WORK!!a.) How many words can you type in 5 minutes?b.) How many words can you type in 8 minutes?c.) How long would it take to read 280 words?
The given function is
[tex]w(x)=70x[/tex]Where x is minutes.
(a) To find the number of words typed in 5 minutes, we just need to replace the variable for 5 and solve
[tex]w(5)=70(5)=350[/tex]Therefore, there are typed 350 words in 5 minutes.
(b) We do the same process for 8 minutes.
[tex]w(8)=70(8)=560[/tex]Therefore, there are typed 560 words in 8 minutes.
(c) To find the type for 280 words, now we replace the other variable w(x), and solve for x
[tex]280=70x[/tex]We divide the equation by 70
[tex]\frac{280}{70}=\frac{70x}{70}\rightarrow x=4[/tex]Therefore, 280 words take 4 minutes.
the sum of 5 times a number and twice its cube
if <5 = 130 and m<3 - 13x. find the value of x Round to the nearest tenth if necessary.
From the figure, it can be seen that ∠3 and ∠5 are co-interior angles. Thus, we find that the value of x is 3.84.
It is given to us that -
∠5 = 130
and, ∠3 = 13x
We have to find out the value of x.
Here, we have two parallel lines. Parallel lines are the lines that do not intersect at any points.
There is also a transversal that passes through the two parallel lines in the same plane intersecting the parallel lines at two separate points.
When a transversal cuts two parallel lines, there are different angles formed as shown in the figure.
∠3 and ∠5 are co-interior angles that lie on the same side of the transversal.
So, ∠3 + ∠5 = 180
=> 13x + 130 = 180
=> 13x = 50
=> x = 3.84
Thus, from the value of the sum of co-interior angles, we find that the value of x is 3.84.
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The scatter plot shows the number of hours worked, x, and the amount of money spent on entertainment, y, by each of 25 students.Use the equation of the line of best fit, =y+1.82x11.36, to answer the questions below.Give exact answers, not rounded approximations. (a) For an increase of one hour in time worked, what is the predicted increase in the amount of money spent on entertainment?$(b) What is the predicted amount of money spent on entertainment for a student who doesn't work any hours?$(c) What is the predicted amount of money spent on entertainment for a student who works 8 hours?$
Solution:
Given the scatterplot below:
where the equation of the line of best fit is expressed as
[tex]y=1.82x+11.36[/tex]A) Predicted increase in the amount of money spent on entertainment, for an increase of one hour in time worked.
Recall that the line equation is expressed as
[tex]\begin{gathered} y=mx+c \\ where \\ m=slope \\ slope=\frac{increase\text{ in y}}{increase\text{ in x}} \end{gathered}[/tex]By comparison with the equation of line of best fit, we see that
[tex]\begin{gathered} slope=1.82 \\ where \\ slope=\frac{increase\text{ in amout of money spent}}{increase\text{ in the number of hours worked}} \end{gathered}[/tex]Thus, we have
[tex]\begin{gathered} 1.82=\frac{increase\text{ in amount of money spent}}{1} \\ \Rightarrow predicted\text{ increase in amount of money spent on entertainment = \$1.82} \end{gathered}[/tex]B) Predicted amount of money spent on entertainment for a student with no number of hours worked
This implies that from the equation of the line of best fit, the value of x is zero.
By substitution, we have
[tex]\begin{gathered} y=1.82(0)+11.36 \\ =0+11.36 \\ \Rightarrow y=\$11.36 \end{gathered}[/tex]C) Predicted amount of money spent on entertainment for a student with8 hours of work.
Thus, we have the value of x to be 8 from the equation of the line of best fit.
By substitution, we have
[tex]\begin{gathered} y=1.82\left(8\right)+11.36 \\ =14.56+11.36 \\ \Rightarrow y=\$25.92 \end{gathered}[/tex]What is the solution of 5|2x + 1| – 3 ≤ 7?
Given
5|2x + 1| – 3 ≤ 7
Find
Solve the inequality
Explanation
[tex]\begin{gathered} 5|2x+1|-3\leq7 \\ 5|2x+1|\leq7+3 \\ 5\lvert2x+1\rvert\leq10 \\ |2x+1|\leq\frac{10}{5} \\ \\ |2x+1|\leq2 \end{gathered}[/tex]we know that
[tex]2x+1\leq2\text{ }and\text{ }2x+1>-2[/tex]so ,
[tex]\begin{gathered} 2x+1\leq2 \\ 2x\leq1 \\ x\leq\frac{1}{2} \\ \\ and \\ \\ 2x+1\ge-2 \\ 2x\ge-2-1 \\ 2x\ge-3 \\ x\ge-\frac{3}{2} \end{gathered}[/tex]so ,
[tex]-\frac{3}{2}\leq x\leq\frac{1}{2}[/tex]Final Answer
Hence , the correct option is
[tex]-\frac{3}{2}\leq x\leq\frac{1}{2}[/tex]5 2/5 × 0.8A. 4.32B. 5.76C.7.80D.2.75
Answer:
A. 4.32
Explanation:
First, we need to transform the mixed number 5 2/5 into a decimal number as:
[tex]5\frac{2}{5}=5+\frac{2}{5}=5+0.4=5.4[/tex]Then, we can multiply 5.4 by 0.8, so:
[tex]5\frac{2}{5}\times0.8=5.4\times0.8=4.32[/tex]To multiply 5.4 by 0.8, we can multiply the numbers normally without taking into account the decimal points. So 54 times 08 is equal to:
Then, 5.4 has one digit after the decimal point and 0.8 has one digit after the decimal point. So, in total, we have two digits after the decimal point. It means that the result is equal to 4.32 because we need two digits after the decimal point.
Therefore, the answer is 4.32
Given R(I, y) = (-y, z) and the point Qt1, 0), what is R(Q)?R(Q)
Given that R(x, y) = (-y, x)
This is a transformation.
We want to find R(Q)
The point Q is given as:
Q = (1, 0)
This means that x = 1 and y = 0
Therefore, for R(Q):
-y = -0 = 0
x = 1
Therefore:
R(Q) = (-y, x) = (0, 1)
I hope you can help me with this I can’t understand it and I’ve already had three tutors turn me down because they didn’t understand it
Given:
An angle whose supplement is 10 degrees more than twice its complement.
Required:
To write and solve the equation.
Explanation:
Let the angle be x degrees.
Supplement of this angle = 180 - x
Complement of this angle = 90 -x
Given that supplement is 10 degrees more than twice its complement.
So the equation becomes:
180- x =2(90 - x) + 10
Solve by multiplication.
180 - x = 180 - 2x +10
Solve by collectiong the like terms.
2x - x = 180 - 180 + 10
x = 10 degrees
Final Answer:
The value of the angle is 10 degrees.
Write a sequence of dilations and transformations that map circle B onto circle A and that shows the two circles you created are similar.
We have start with circle B, which is a circle with radius equals to 3 and centered at (4, 4).
The circle A has a radius equals to 4. If we dilate the image of the circle B around the center of circle A by the ratio between the radius, we're going to have circle A.
Translation means the displacement of a figure or a shape from one place to another. If we translate the circle B 8 units to the left, The image is going to have the same center as circle A.
The transformations that takes circle B to circle A are a dilation around (4, 4) by a factor of 4/3, and then a horizontal translation of 8 units to the left.
Order these numbers from least to greatest 7.15 , 7 18/25 , 7.134 , 77/10
If a || band e l f, what is the value of y?(x + 1)[(x-3°
y = x + 1 [ alternate exterior angles ]
I'm a bit confused and I'm not sure but I think it's D?
125x + 200 ≥ 1,200
Solve for x
x ≥ (1200 - 200)/125
x ≥ 1000/125
x≥ 8
Then graph solution is
Arrow to right, beggining at 8, including 8 point
Answer is OPTION A)
Find the missing sides of the triangle. Leave youranswers as simplified radicals.
Consider the following right triangle:
To find the missing sides x and y, we can apply the following trigonometric ratios:
[tex]\cos(60^{\circ})=\frac{adjacent\text{ side to the angle 60}^{\circ}}{Hypotenuse}[/tex][tex]\sin(60^{\circ})=\frac{opposite\text{ side to the angle 60}^{\circ}}{Hypotenuse}[/tex]and
[tex]\tan(60^{\circ})=\frac{opposite\text{ side to the angle 60}^{\circ}}{adjacent\text{ side to the angle 60}^{\circ}}[/tex]thus, applying the data of the problem to the last equation, we get:
[tex]\tan(60^{\circ})=\frac{opposite\text{ side to the angle 60}^{\circ}}{adjacent\text{ side to the angle 60}^{\circ}}=\frac{15}{y}[/tex]that is:
[tex]\tan(60^{\circ})=\frac{15}{y}[/tex]solving for y, we obtain:
[tex]y=\frac{15}{\tan(60^{\circ})}=\frac{15}{\sqrt{3}}[/tex]On the other hand, applying the above data to the first equation, we get:
[tex]\cos(60^{\circ})=\frac{adjacent\text{ side to the angle 60}^{\circ}}{Hypotenuse}=\frac{y}{x}=\frac{15}{\sqrt{3}}\text{ }\cdot\frac{1}{x}[/tex]or
[tex]\cos(60^{\circ})=\frac{15}{\sqrt{3}\text{ x}}\text{ }[/tex]solving for x, we obtain:
[tex]x=\frac{15}{\sqrt{3}\cdot\cos(60)}=\text{ }\frac{15}{\sqrt{3\text{ }}\cdot1/2}=\frac{2(15)}{\sqrt{3}}=\frac{30}{\sqrt{3}}[/tex]we can conclude that the correct answer is:
Answer:[tex]x=\frac{30}{\sqrt{3}}[/tex]and
[tex]y=\frac{15}{\sqrt{3}}[/tex]