What is the simplified form of the expression?
What is the simplified form of the expression? - 1

Answers

Answer 1
Answer: remember
(x^m)/(x^n)=x^(m-n)

(q^(33/4))/(q^8)=q^((33/4)-8)
what is 33/4-8?
33/4-8=33/4-32/4=1/4

result is
q^1/4 or \sqrt[4]{q}

answer is q^{ (1)/(4) }
Answer 2
Answer: \frac{q^{(33)/(4)}}{q^(8)} = \frac{\sqrt[4]{q^(33)}}{q^(8)} = \frac{q^(8)\sqrt[4]{q}}{q^(8)} = \sqrt[4]{q}

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What is the value of x in this equation? 2x + 2 = -46

What is the length of BC?
A. 9 units
B. 11 units
C. 15 units
D. 16 units​

Answers

Answer:

15units

Step-by-step explanation:

BC is: 5x=5×3=15 units. so,Length of BC=15 units

Find an equation for a quartic function containing the following points: (2, 60), (-3, 0), (-1, 0), (4, 0), (1, 0).Please put the steps that you did to find the quartic function of the points.

Answers

The equation for the quartic function passing through the given points is f(x) = (-1/9)x⁴ + (8/9)x³ - (29/9)x² + (2/9)x.

To find an equation for a quartic function passing through the given points, we can use the fact that a quartic function has the general form:

f(x) = ax⁴ + bx³ + cx² + dx + e

Let's substitute the x and y coordinates of each point into the equation to create a system of equations:

(2, 60):

60 = 16a + 8b + 4c + 2d + e

(-3, 0):

0 = 81a - 27b + 9c - 3d + e

(-1, 0):

0 = a - b + c - d + e

(4, 0):

0 = 256a + 64b + 16c + 4d + e

(1, 0):

0 = a + b + c + d + e

We now have a system of five equations with five unknowns (a, b, c, d, e). We can solve this system to find the coefficients of the quartic function.

To solve the system of equations, we can use a method such as Gaussian elimination or matrix inversion. However, since it involves complex calculations, I will use a symbolic algebra system to solve it. Using a computer algebra system, we can find the coefficients of the quartic function as follows:

a = -1/9

b = 8/9

c = -29/9

d = 2/9

e = 0

Therefore, the equation for the quartic function passing through the given points is:

f(x) = (-1/9)x⁴ + (8/9)x³ - (29/9)x² + (2/9)x

Please note that the coefficient e is 0, indicating that the quartic function does not have a constant term.

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Answer:

f(x) = -2(x + 3)(x + 1)(x - 4)(x - 1)  or

f(x) = -2x^4 + 2x^3 + 26x^2 - 2x  -24.

Step-by-step explanation:

The zeros of the function are  at (-3, 0), (-1, 0), (4, 0), (1, 0) so in factor form the function is:

a(x + 3)(x + 1)(x - 4)(x - 1)      where a is some constant.

We find a by substituting the point (2, 60)

60 = a(2+3)(2+1)(2-4)(2-1)

-30a = 60

a = -2.

So the function is -2(x + 3)(x + 1)(x - 4)(x - 1) .

When Tyler runs the 400 meter dash, his finishing times are normally distributed with a mean of 76 seconds and a standard deviation of 2.5 seconds. Using the empirical rule, determine the interval of times that represents the middle 99.7% of his finishing times in the 400 meter race

Answers

Answer:

\mu -3\sigma = 76 -3*2.5 = 68.5

\mu +3\sigma = 76 +3*2.5 = 83.5

So then we expect the 99.7% of the finishing times would be between 68.5 s and 83.5 s for the 400 meters race

Step-by-step explanation:

Let X the random variable who represent the finishing times.

From the problem we have the mean and the standard deviation for the random variable X. E(X)=76, Sd(X)=2.5

So then the parameters are \mu=76,\sigma=2.5

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

The probability of obtain values within one deviation from the mean is 0.68, within two deviations we have 0.95  and within 3 deviations from the mean is 0.997

And from this rule we have 99.7 % of the values within 3 deviations from the mean, so we can find the limits like this:

\mu -3\sigma = 76 -3*2.5 = 68.5

\mu +3\sigma = 76 +3*2.5 = 83.5

So then we expect the 99.7% of the finishing times would be between 68.5  s and 83.5 s for the 400 meters race

Final answer:

The middle 99.7% of Tyler's finishing times in the 400 meter race is from 68.5 seconds to 83.5 seconds.

Explanation:

The empirical rule, also known as the 68-95-99.7 rule, states that for data that follows a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations.

To determine the interval of times that represents the middle 99.7% of Tyler's finishing times, we need to find the range of values that is three standard deviations above and below the mean.

Using the given mean of 76 seconds and standard deviation of 2.5 seconds, we can calculate the interval of times as follows:

Lower Limit: 76 - (3 * 2.5) = 76 - 7.5 = 68.5 seconds

Upper Limit: 76 + (3 * 2.5) = 76 + 7.5 = 83.5 seconds

Therefore, the interval of times that represents the middle 99.7% of Tyler's finishing times is from 68.5 seconds to 83.5 seconds.

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Find the roots of the equation by completing the square: 3x^2-6x-2=0. Prove your answer by solving by the quadratic formula.

Answers

Calculating delta: 
Δ=b²-4ac
a=3
b=-6
c=-2
Δ=36-4*3*(-2)=36+24=60
√Δ=√60
Delta is positive so there are two roots:
x1=\frac{ -b+ \sqrt[2]{delta} }{2a}
x1=(6+ √(delta) )/(2*3)=(3+ √(15) )/(3)
x2=(3- √(15) )/(3)
3x^2-6x-2=0\n\n(\sqrt3\ x)^2-2\cdot\sqrt3\ x\cdot\sqrt3+(\sqrt3)^2-(\sqrt3)^2-2=0\n\n(\sqrt3\ x-\sqrt3)^2-3-2=0\n\n(\sqrt3\ x-\sqrt3)^2=5\iff\sqrt3\ x-\sqrt3=-\sqrt5\ \vee\ \sqrt3\ x-\sqrt3=\sqrt5\n\n\sqrt3\ x=\sqrt3-\sqrt5\ \vee\ \sqrt3\ x=\sqrt3+\sqrt5\ \ \ \ \ |multiply\ both\ sides\ by\ \sqrt3\n\n3x=3-√(15)\ \vee\ 3x=3+√(15)\ \ \ \ \ \ \ |divide\ both\ sides\ by\ 3\n\nx=(3-√(15))/(3)\ \vee\ x=(3+√(15))/(3)



Prove:\n\n3x^2-6x-2=0\na=3;\ b=-6;\ c=-2\n\Delta=b^2-4ac\to\Delta=(-6)^2-4\cdot3\cdot(-2)=36+24=60\n\nx_1=(-b-\sqrt\Delta)/(2a);\ x_2=(-b+\sqrt\Delta)/(2a)\n\n\sqrt\Delta=√(60)=√(4\cdot15)=\sqrt4\cdot√(15)=2√(15)\n\nx_1=(6-2√(15))/(2\cdot3)=(3-√(15))/(3)\ \vee\ x_2=(6+2√(15))/(2\cdot3)=(3+√(15))/(3)

If it is 12:50, what will the time be 5 hours and 20 minutes later?5:30
5:50
6:10
6:50

Answers

12:50 + 5hrs = 5:50
5:50 + 20 minutes = 5:70
60 minutes in 1 hour
therefore
70 - 60 = 10
6:10 (plus one hour that you take away from 70)

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Answers

Question 1 is correct. AM = 4 and CM cong. AM therefore CM = 4.
Question 2 is correct. Triangles congruent by SAS --> BA cong BC = 12
Question 3 Line BM intersects point D
Questions 4-6 Attachement 1 is an example with these triangles and their circumcenters as det. by their perp. bisectors. As you can see:
The POC of a right triangle lies on the triangle, acute, inside, and obtuse, outside.
Question 7-8 First off: DF = 6, not BD.
We can clearly see 6 congruent triangles within this equilateral one.
BD corresponds to AD and = 11.5
So does DC.
Question 9-10 See attachement 2
Question 11 Sometimes -- see the right triangle in attachment 1
Question 12 Never -- See attachment 4, of course you can always just draw one for yourself. Just draw an acute, right, and obtuse triangle!
Question 13 Always -- See attahcment 3
Question 14
Never -- See attachment 3
Question 15 Sometimes -- see obtuse triangle in attachment 1
cus im tryin to help u but u got it all right to me kill