The degree 3 polynomial is "x³-2x²-64x+128".
We have to find a polynomial of degree 3.We are given that the polynomial has -8, 2, and 8 as its roots.We are also given that the coefficient of x³ equals 1.If x = -8 is a root, then:(x+8) is a factor of the polynomial.If x = 2 is a root, then:(x-2) is a factor of the polynomial.If x = 8 is a root, then:(x-8) is a factor of the polynomial.All the three roots are of the same polynomial.So, the polynomial is the product of these factors.(x+8)(x-8)(x-2)(x²-64)(x-2)x³-2x²-64x+128The required polynomial is "x³-2x²-64x+128".To learn more about polynomials, visit :
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3 1/3(y+1 1/8)=6 11/12
Answer:
(3 1/3)·(y + (1 1/8)) = 6 11/12
(9/3 + 1/3)·(y + (8/8 + 1/8)) = 72/12 + 11/12
10/3·(y + 9/8) = 83/12
y + 9/8 = 83/12·3/10
y + 9/8 = 83/4·1/10
y + 9/8 = 83/40
y = 83/40 - 9/8
y = 83/40 - 45/40
y = 38/40
y = 19/20
Graph the line with slope
-2/3
passing through the point (2, -3).
The solution is written down, and the graph is drawb below
Given g(x) = 1/x^3Explain if the question cannot be solved
Given
[tex]g(x)=\frac{1}{x^3}[/tex]To find:
[tex]\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}g(x)dx[/tex]Explanation:
It is given that,
[tex]g(x)=\frac{1}{x^3}[/tex]That implies,
[tex]\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}g(x)dx[/tex]Evaluate the correlation shown in this scatter plot and then answer the 2 questions below.
How would you describe the direction and strength of this scatter plot? Is it positive or negative? Is it weak, moderately strong, or perfect? (worth 1.5 points)
How did you decide what words to choose to describe this correlation? (worth 1.5 points) 30 POINTS FORR WHO AWNSERS
The given scatter plot points are increasing, indicating a rise in data points, direction oriented to the right and the strength of scatter plot points correlation is moderately strong.
A graph with dots is shown to indicate the relationship between two sets of data.
According to the given scatter plot, the scatter plot points are increasing, indicating a rise in data points, and we may conclude that the correlation is positive.
The scatter plots are now oriented to the right. As a result, we may claim that the correlation is moderately strong.
Thus. the given scatter plot points are increasing, indicating a rise in data points, direction oriented to the right and the strength of scatter plot points correlation is moderately strong.
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Write and graph a direct variation equation that passes through the given point.(2, 5)Write the direct variation equation.y=
For direct variation, an increase or decrease in one variable leads to a proportional increase or decrease in the other variable
Given that y varies directly with x, we would introduce a constant of proportionality, k
The expresion becomes
y = kx
For x = 2 and y = 5,
5 = 2k
k = 5/2 = 2.5
The direct variation equation is
y = 2.5x
Hello! I can’t seem to remember to do this answer
Given the following inequality:
[tex]2x\leq10[/tex]We will solve the inequality as follows:
Divide both sides by 2
[tex]\begin{gathered} \frac{2x}{2}\leq\frac{10}{2} \\ \\ x\leq5 \end{gathered}[/tex]So, the solution to the inequality x = (-∞, 5]
So, x = 5 is a solution to the inequality
So, the answer will be yes; x = 5 is a solution.
answer f 1 half 25 y intercept equals 375--g slope 1 half 25 y intercept equal 15H slope equals 25 y intercept equal 375J slope equals negative 25 y intercept equals 15
Answer:
[tex]\begin{gathered} \text{Slope}=-\frac{1}{25} \\ y-\text{intercept}=15 \end{gathered}[/tex]Step-by-step explanation:
Linear functions are represented by the following expression:
[tex]\begin{gathered} y=mx+b \\ \text{where,} \\ m=\text{slope} \\ b=y-\text{intercept} \end{gathered}[/tex]m is the constant rate of change of the function, and it's calculated as the change in y over the change in x:
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{14.6-15}{10-0} \\ m=-\frac{1}{25} \end{gathered}[/tex]The y-intercept of a linear function is when the line crosses the y-axis, which means when x=0.
Therefore, the y-intercept of the line is 15.
In July 1 2020 culver inc invested $635250 in a mine estimated to have 847000 tons of ore of uniform grade during the last 6 months of 2020 146000 tons of ore were mined calculate depletion cost per unit
Given:
culver inc invested $635250 in a mine estimated to have 847000 tons of ore of uniform grade
So, the estimated cost per unit =
[tex]\frac{635250}{847000}=0.75[/tex]This means the cost = $0.75 per ton
A random sample of 860 births in a state included 423 boys. Construct a 95%
confidence interval estimate of the proportion of boys in all births. It is believed that
among all births, the proportion of boys is 0.513. Do these sample results provide
strong evidence against that belief?
Construct a 95% confidence interval estimate of the proportion of boys in all births.
Using the z-distribution, it is found that the 95% confidence interval is (0.45 , 0.52), and it does not provide strong evidence against that belief.
A confidence interval of proportions is given by:
[tex]\pi[/tex] ± [tex]z\sqrt{\frac{\pi (1-\pi )}{n} }[/tex]
where [tex]\pi[/tex] is the sample proportion, z is the critical value and n is the sample size.
In this problem, we have 95% confidence level, hence [tex]\alpha[/tex] = 0.95, z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2}[/tex] = 0.975, so the critical value is z = 1.96
We have that a random sample of 860 births in a state included 423 boys, hence the parameters are given by:
n = 864, [tex]\pi =\frac{423}{860}[/tex] = 0.49
Then the bounds of the interval are given by:
[tex]\pi[/tex] + [tex]z\sqrt{\frac{\pi (1-\pi )}{n} }[/tex] = 0.49 + [tex]1.96\sqrt{\frac{0.49(0.513)}{860} }[/tex] = 0.52
[tex]\pi[/tex] - [tex]z\sqrt{\frac{\pi (1-\pi )}{n} }[/tex] = 0.49 - [tex]1.96\sqrt{\frac{0.49(0.513)}{860} }[/tex] = 0.45
The 95% confidence interval estimate of the population of boys in all births is (0.45 , 0.52). Since the interval contains 0.513, it does not provide strong evidence against that belief.
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Find the midpoint of the segment with endpoints of (-1, 7) and (3,-3) andenter its coordinates as an ordered pair. If necessary, express coordinates asfractions, using the slash mark (/) for the fraction bar.
We are given the points (-1,7) and (3,-3) and we want to calculate the midpoint of the segment that joins this points. REcall that given points (a,b) and (c,d), the midpoint of the segment that joins the points is calculated by averaging each coordinate (that is,adding the coordinates and then dividing them by 2). So the midpoint would be
[tex](\frac{a+c}{2},\frac{b+d}{2})[/tex]you and your family are taking a trip to Brazil. You are bringing $175 on the trip. The rate pf currency exchange is 4.65 Real (Brazilian money) per 1 United States dollar. How many Real will you have on the trip?
Answer:
813.75 can you have .75 of a real? if not, then 813
Step-by-step explanation:
175 x 4.65 = 813.75
14) Solve the following quadratic equations by using the quadratic formula a) 3x2 - 7x + 4 = 0 b) 5x2 + 3x = 9
Quadratic formula
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Where a is the coefficient of the first term, b the coefficient of the second term and c the coefficient of third term
a)
[tex]3x^2-7x+4=0[/tex]replacing on the quadratic formula
[tex]x=\frac{-(-7)\pm\sqrt[]{(-7)^2-4(3)(4)}}{2(3)}[/tex]simplify
[tex]\begin{gathered} x=\frac{7\pm\sqrt[]{49-48}}{6} \\ \\ x=\frac{7\pm\sqrt[]{1}}{6} \\ \\ x=\frac{7\pm1}{6} \end{gathered}[/tex]x has two solutions
[tex]\begin{gathered} x_1=\frac{7+1}{6}=\frac{4}{3} \\ \\ x_2=\frac{7-1}{6}=1 \end{gathered}[/tex]b)
[tex]5x^2+3x=9[/tex]rewrite on general form
[tex]5x^2+3x-9=0[/tex]raplace on quadratic formula
[tex]x=\frac{-(3)\pm\sqrt[]{(3)^2-4(5)(-9)}}{2(5)}[/tex]simplify
[tex]\begin{gathered} x=\frac{-3\pm\sqrt[]{9+180}}{10} \\ \\ x=\frac{-3\pm\sqrt[]{189}}{10} \end{gathered}[/tex]x has two solutions
[tex]\begin{gathered} x_1=\frac{-3+\sqrt[]{189}}{10}\approx1.075 \\ \\ x_2=\frac{-3-\sqrt[]{189}}{10}\approx-1.67 \end{gathered}[/tex]Skills Find the new bank account balance Old balance: $500.00Withdrawal: $175.00Withdrawal: $60.00Deposit: $37.50
In order to find the new bank account balance, we can do a sum of all the deposits and subtraction of all withdrawals to the old balance account,
[tex]\begin{gathered} BA=500.00-175.00-60.00+37.50 \\ BA=302.50 \end{gathered}[/tex]#12 Suppose Bob is a 9th grade student. Bob's Mom plans to buy him a VA529 education bondof $50,000 education bond for his college education. If the discount rate is 6%, compounded semi-annually, what price should Bob's Mom pay now (current price)? (Hint: there are 4 years from 9thgrade to college.)
For solving this exercise, we will use the formula of the Present or Current value, as follows:
[tex]PV\text{ = FV }\ast\text{ }\lbrack\frac{1}{\square}(1+i)^n\rbrack[/tex]Where:
PV = Present or Current Value
FV = Future Value
i = Interest or discount rate
n = Periods of time
Replacing with the values we know:
[tex]PV\text{ = 50,000 }\ast\text{ }\lbrack\text{ 1 }\frac{\square}{\square}\text{ }(1+0.03)^8\rbrack[/tex]PV = 50,000 * 0.7894
PV = 39,470
Bob's Mom
how to solve 4|x|+|-4|=|-6|
x = 1/2, x = -1/2
Simplify:
4|x| + |-4| = |-6|
4|x| + 4 = 6
4|x| = 2
|x| = 1/2
Solutions:
1) x = 1/2
2) x = -1/2
Please help I was sick today and I don’t understand
Answer:
4
Step-by-step explanation:
By the exterior angle theorem,
[tex]27x+2=65+10x+5 \\ \\ 27x+2=10x+70 \\ \\ 17x=68 \\ \\ x=4[/tex]
In New York the mean salary for high school teachers in 2017 was 97010 with a standard deviation of 9540. Only Alaska’s mean salary was higher. Assume new York’s state salaries follow a normal distribution. (A) what percent of new York’s high school teachers earn between 83,000 and 88,000? (B) what percent of New York teachers earn between 88,000 and 103,000?
(C) what percent of new York’s state high school teachers earn less than 73,000?
a. 20.31% of New York's high school teachers earn between 83,000 and 88,000
b. 18.49% of New York teachers earn between 88,000 and 103,000
c. 1.19% of New York’s state high school teachers earn less than 73,000
Given,
The salary for high school teachers in 2017 = 97010
Standard deviation = 9540
Consider salaries as normal distribution.
Here,
Mean, μ = 97010, Standard deviation, σ = 9540
a. Percentage of New York's high school teachers earn between 83,000 and 88,000
The proportion is the p-value of Z when X = 88,000 subtracted by the p-value of Z when X = 83,000.
That is,
X = 88,000
Z = (X - μ) / σ = (88,000 - 97010) / 9540 = -9010/9540 = -0.944
The p value of z score - 0.944 is 0.3452
Next,
X = 83,000
Z = (X - μ) / σ = (83,000 - 97010) / 9540 = -14010/9540 = -1.468
The p value of z score - 1.468 is 0.1421
Then,
0.3452 - 0.1421 = 0.2031 = 20.31%
That is,
20.31% of New York's high school teachers earn between 83,000 and 88,000
b. Percentage of New York teachers earn between 88,000 and 103,000
The proportion is the p-value of Z when X = 103,000 subtracted by the p-value of Z when X = 88,000
X = 103,000
Z = (X - μ) / σ = (103,000 - 97010) / 9540 = 5990/9540 = 0.6279
The p value of z score 0.6279 is 0.5301
Next,
X = 88,000
Z = (X - μ) / σ = (88,000 - 97010) / 9540 = -9010/9540 = -0.944
The p value of z score - 0.944 is 0.3452
Then,
0.5301 - 0.3452 = 0.1849 = 18.49%
That is,
18.49% of New York teachers earn between 88,000 and 103,000
c. Percentage of new York’s state high school teachers earn less than 73,000
The proportion is the p-value of Z when X = 73000
X = 73,000
Z = (X - μ) / σ = (73,000 - 97010) / 9540 = -24010/9540 = -2.516
The p value of z score - 2.516 is 0.0119
That is,
1.19% of New York’s state high school teachers earn less than 73,000
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Let theta equals 11 times pi over 12 periodPart A: Determine tan θ using the sum formula. Show all necessary work in the calculation.Part B: Determine cos θ using the difference formula. Show all necessary work in the calculation.
The fisrt part is the divide your angle into two angles, could be 6/12π and 5/12π
[tex]\begin{gathered} A=\frac{4\pi}{12}=\frac{\pi}{3} \\ B=\frac{7\pi}{12} \end{gathered}[/tex]For the sum formula:
[tex]\begin{gathered} \tan (\theta)=\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\cdot\tan B} \\ \tan (A+B)=\frac{1.73-3.73}{1-1.73\cdot(-3.73)} \\ \tan (A+B)=\frac{-2}{7.45}=-0.27 \end{gathered}[/tex]For the difference formula:
[tex]\begin{gathered} A=\frac{1\pi}{12} \\ B=\pi \end{gathered}[/tex][tex]\begin{gathered} \tan (B-A)=\frac{\tan B-\tan A}{1+\tan A\cdot\tan B} \\ \tan (B-A)=\frac{0-0.268}{1+0\cdot0.267} \\ \tan (B-A)=-0.268 \end{gathered}[/tex]Both methods work and result in the same answeer
Find the additive inverse. −31
Answer:
To get the additive inverse of a positive number you put a minus in front of it and to get the additive inverse of a negative number, you remove the minus to make it a positive number.
Write (2p^2)^3 without exponents.(2p^2)^3 = ??
We are required to write the expression:
[tex](2p^2)^3[/tex]Without exponents. First, we operate the parentheses:
[tex](2p^2)^3=2^3(p^2)^3=2^3p^6[/tex]This is the simplified expression. If we wanted to avoid the exponents, then we have to express the exponents as products:
[tex]2^3p^6=2\cdot2\cdot2\cdot p\cdot p\cdot p\cdot p\cdot p\cdot p[/tex]This is the required expression
A chef at a restaurant uses 19 pounds of butter each day. About how many grams of butter does the chef use each day? Use the conversion factors
28.4 grams
1 ounce
16 ounces
1 pound
and
Using the conversion factors, 454 grams of butter is used by the chef each day.
What are conversion factors?A conversion factor is a number that is used to multiply or divide one set of units into another. If a conversion is required, it must be done using the correct conversion factor to get an equal value. For instance, 12 inches equals one foot when converting between inches and feet.So, grams of butter used each day:
19 pounds = 8618.26 gramsThen,
1 pound = 8618.26/191 pounds = 453.592 gramsRounding off: 454 gramsTherefore, using the conversion factors, 454 grams of butter is used by the chef each day.
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The correct question is given below:
A chef at a restaurant uses 19 pounds of butter each day. About how many grams of butter does the chef use each day? Use the conversion factors
8 if x ≤-1
2x if -1 < x <4
-4 - x + 6 if x ≥ 4)
2x=-1+6
2x=5
x=3
so the answer is 3
You paid $600 for a new guitar. Your guitar cost $40 more than twice the cost of your friends guitar. Wright an equation based on this information.
Answer: 600 divided by 2 + 40
Step-by-step explanation:
Answer:
2x + 40 = 600
2x = 560
x = 280.
Your friends guitar costs $280.
Step-by-step explanation:
Set the equation equal to $600, the cost of the new guitar. Let the variable, x, represent the cost of your friends guitar. Since your guitar was + 40 more than double the cost of your friend’s, this can be written as an equation:
2x + 40 = 600
$150,000 is still owed on a home loan. The loan has an interest rate of 4.5% compounded monthly and calls for monthly payments. The current payment is $1,100 per month. How long will it take to pay the loan off? R
Based on the monthly payment of $1,100, it will take 16 years to liquidate the loan.
How is the number of periods determined?The number of periods is a function of the total payments that will be paid, given the compounding of the interest at 4.5% and the monthly payment of $1,100.
Using an online finance calculator, the total number of years is determined as follows:
I/Y (Interest per year) = 4.5%
PV (Present Value) = $150,000
PMT (Periodic Payment) = $-1,100
FV (Future Value) = $0
Results:
N (# of periods) = 191.328 months
= 16 years (191.328/12)
Sum of all periodic payments = $-210,460.39 (191.328 x $1,100)
Total Interest = $60,460.39
Thus, before the loan is paid off, it will take 16 years or 191 monthly payments.
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Line BC Is a tangent to circle A at Point B. How would I find the measure of angle BCA? I need more explanation
SOLUTION
Notice that line BA is a radius of the circle.
Since line BC is a tangen then the measure of angle ABC is:
[tex]m\angle ABC=90^{\circ}[/tex]Using Triangle Angle-Sum Theorem, it follows:
[tex]m\angle ABC+m\angle BAC+m\angle BCA=180^{\circ}[/tex]This gives:
[tex]90^{\circ}+57^{\circ}+m\angle BCA=180^{\circ}[/tex]Solving the equation gives:
[tex]\begin{gathered} 147^{\circ}+m\angle BCA=180^{\circ} \\ m\angle BCA=180^{\circ}-147^{\circ} \\ m\angle BCA=33^{\circ} \end{gathered}[/tex]Therefore the required answer is:
[tex]m\angle BCA=33^{\circ}[/tex]which best describes the relationship between the two lines described below?
If two lines are perpendicular, then the product of their slopes is equal to -1.
If two lines are parallel, then their slopes are equal.
Write the equation of the lines P and Q in slope-intercept form by isolating y. Compare their slopes to see if they are either parallel or pependicular, or none.
The equation of a line with slope m and y-intercept b in slope-intercept form, is:
[tex]y=mx+b[/tex]Line P:
[tex]\begin{gathered} 6x+3y=12 \\ \Rightarrow3y=-6x+12 \\ \Rightarrow y=\frac{-6x+12}{3} \\ \therefore y=-2x+4 \end{gathered}[/tex]Then, the slope of the line P is -2.
Line Q:
[tex]\begin{gathered} -4x=2y-2 \\ \Rightarrow2y=-4x+2 \\ \Rightarrow y=\frac{-4x+2}{2} \\ \Rightarrow y=-2x+1 \end{gathered}[/tex]Then, the slope of the line Q is -2.
Since both lines have the same slope, then they are parallel.
What is the slope of the line whose equation is y-4=(x-2)?6543qu6543+2w9(0,-1)(2,4)2 3 4 5 6 x
The point-slope form of the equation of a line is given to be:
[tex]y-y_1=m(x-x_1)[/tex]where
[tex]\begin{gathered} m=slope \\ (x_1,y_1)=point\text{ }on\text{ }the\text{ }line \end{gathered}[/tex]The equation of the line is given to be:
[tex]y-4=\frac{5}{2}(x-2)[/tex]This means that the slope is 5/2.
Given P(A) = 0.5, P(B) = 0.65 and P(AUB) = 0.75,find P(ANB).
P(A∪B) = P(A) + P(B) - P(A∩B)
where P(A) is the probability of A happening
P(B) is the probability of B happening
P(A∪B) is the probability of A or B happening
P(A∩B) is the probability of A and B happening
P(A) = 0.5, P(B) = 0.65 and P(AUB) = 0.75
.75 = .5+ .65 - P(A∩B)
.75 =1.15 - P(A∩B)
.75 - 1.15 = -P(A∩B)
-.4 = -P(A∩B)
.4 =P(A∩B)
P(A∩B) = .4
in point a the diameter is 12 C D equals 8 and MCD equals 90 find a measure round to the nearest hundred
Given that BE is diameter and CD is perpendicular, we can deduct that arcs CE and DE are equal.
[tex]\begin{gathered} arc(CD)=arc(CE)+arc(DE) \\ 90=2\cdot arc(CE) \\ \text{arc(CE)}=45=arc(DE) \end{gathered}[/tex]Therefore, arc DE is 45°.
How can you use the Power of a Quotient, Quotient of Powers, Zero Exponent Laws Identity Exponent and to evaluate numerical expressions with whole-number exponents?
Though working with laws of exponents, when part of the exponential equations with the same base, the exponent need to be subtracted.
Since the Quotient of Powers rules is utilized to decrease the number of terms when parts like terms with exponents. In arrange to urge the solution, fair remove the exponents whereas dividing the terms with the same base is appropriate for exponents with the same bases. when powers are multiplied, for two whole numbers of the same bases the exponents are added, while when powers are divided for two whole numbers of the same bases, exponents are subtracted. When any number is raised to the power of zero it comes about in 1 is the zero exponent rule.
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