Find the product.<br> (2y - 2nb)(y + 4nb)<br> Enter the correct answer.

Simplify the product.<br> ((3x-6)/(x+5))·((2x+10)/(x-5))

cricket20 [7]

I hope this helps you

3 0

3 years ago

The number of chaperones needed for a school field trip was s directly proportional to the number of students going on a field t

Korvikt [17]

Answer:

The correct answer is 16.

Step-by-step explanation:

Number of chaperones required for 96 students is 12.

Thus under each chaperone there are $\frac{96}{12}$ = 8 students.

Now a total of 125 students are supposed to be sent on a trip.

Therefore the total number of chaperones required for these 125 students are $\frac{125}{8} = 15.625$ ≈ 16. (converted to the highest integer because the leftover 5 students would need one more chaperones)

Therefore a total for 16 chaperones are required by the Wachusett Perineal School to take 125 students for a field trip.

4 0

3 years ago

2^X = 17^X<br><br> How do you solve that equation?

Kobotan [32]

For this case we must solve the following equation:

$2 ^ x = 17 ^ x$

For this, we follow the steps below:

We find the Neperian logarithm on both sides of the equation to remove the variable from the exponent

$ln (2 ^ x) = ln (17 ^ x)$

We use one of the logarithm properties to extract x from the exponent:

$xln (2) = xln (7)$

We subtract xln (7) on both sides of the equation:

$xln (2) -xln (7) = 0$

We take x common factor:

$x (ln (2) -ln (7)) = 0$

We divide between $(ln (2) -ln (7))$ on both sides of the equation, then:

$x = 0$

Answer:

$x = 0$

5 0

3 years ago

Is the polynomial prime? <br> 7x2-35x+2x-10

wolverine [178]

The given polynomial is not a prime polynomial.

Given the polynomials, we have to choose the polynomial which is prime.

A polynomial with integer coefficients that cannot be factored into lower degree polynomials is Prime polynomials.

$7x^2-35x+2x-10$

$7x^2-33x-10$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

<h3>What is the formula for the quadratic roots?</h3>

$\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=7,\:b=-33,\:c=-10$

$x_{1,\:2}=\frac{-\left(-33\right)\pm \sqrt{\left(-33\right)^2-4\cdot \:7\left(-10\right)}}{2\cdot \:7}$

$x_{1,\:2}=\frac{-\left(-33\right)\pm \:37}{2\cdot \:7}$

$x_1=\frac{-\left(-33\right)+37}{2\cdot \:7},\:x_2=\frac{-\left(-33\right)-37}{2\cdot \:7}$

$x=5,\:x=-\frac{2}{7}$

Therefore the factor is (x-5) and (7x+2).

can be factored

∴ not a prime polynomial.

To learn more about the prime polynomial visit:

brainly.com/question/26388060

#SPJ1

4 0

1 year ago

HELP

alina1380 [7]

Answer:

D

Step-by-step explanation:

72+21-11=82

82-67=15

3 0

2 years ago

Read 2 more answers

Find the product. (2y - 2nb)(y + 4nb) Enter the correct answer.