Multiply (5x2 x − 4)(x 2). 5x3 x2 − 4x − 8 5x3 11x2 − 2x − 8 5x3 10x2 − 4x − 8 5x3 20x2 2x − 8

Answers

Answer 1

Answer: (5x^2 + x - 4)(x + 2) = x(5x^2 + x - 4) + 2(5x^2 + x - 4) = 5x^3 + x^2 - 4x + 10x^2 + 2x - 8 = 5x^3 + 11x^2 - 2x - 8

Answer 2

Answer:

Answer:

5x^3 + 11x^2 - 2x - 8

Step-by-step explanation:

took the test hope this helps please mark brainlisets

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Which quadratic function has a leading coefficient of 2 and a constant term of –3?f(x) = 2x^3 – 3

f(x) = –3x^2 – 3x + 2

f(x) = –3x^3 + 2

f(x) = 2x^2 + 3x – 3

Answers

quadatic means 2nd degree, means highest exponent of placeholder is 2
leading coefient is the number that is in front of the highest power x
constant term is the one that doesn't have an x witht it

quadratic
2nd and last fn are quadratic

leading coefient of 2 is first and last fn

constant term is -3 for first and last fn

answer is last function

In a school 3/5are boys. In a day 1/6 were absent and 250 boys were present. How many girls are in that school

Answers

Answer:

There are 200 girls in that school

Step-by-step explanation:

The correct and complete question is as folly;

In a school 3/5 pupils are boys. One day 1/6 of the boys were absent when 250 boys were present. How many girls are in the school?

SOLUTION

Let the total number of students in the school be x students

Since 3/5 are boys , then the number of girls in the school would be 1-3/5 = 2/5

The number of boys are 3/5 * x = 3x/5

The number of girls are 2/5 * x = 2x/5

Now on a particular day, 1/6 of the boys were absent and 250 boys were present.

What this means is that the fraction of boys present is 1-1/6 = 5/6

Now, 5/6 of the total boys population were present.

Mathematically;

5/6 * 3x/5 = 250

3x/6 = 250

x/2 = 250

x = 2 * 250 = 500

So there are 590 students in the school.

The number of girls in the school is ;

2x/5 = 2/5 * 500 = 200 girls

Simplify the expression 6^2/^3

Answers

the expression is
20/3

Select the equation of the line that passes through the point (2, 6) and is perpendicular to the line x = 4.y = 4

y = 6

x = 6

x = 2

Answers

Hello,
All the perpendiculars of the line x=4 are parallele to the oy axis.

y=6
Answer B

The equation of the perpendicular line is y = 6.

For the graphed exponential equation, calculate the average rate of change from x = −3 to x = 0. graph of f of x equals 0.5 to the x power, minus 6.

Answers

The average rate of change will be negative 7 / 3.

How to find the average rate of change of something?

Let the thing that is changing be y and the thing with which the rate is being compared is x, then we have the average rate of change of y as x changes as:

Average rate = (y₂ - y₁) / (x₂ - x₁)

The exponential function is given below.

f(x) = (0.5)ˣ - 6

Then the average rate of change will be

Average rate = [f(0) - f(-3)] / [0 - (-3)]

Average rate = [(0.5⁰ - 6) - (0.5⁻³ - 6)] / 3

Average rate = [1 - 6 - 8 + 6] / 3

Average rate = - 7 / 3

More about the average rate of change link is given below.

brainly.com/question/23715190

#SPJ2

I hope that this helps :) :) :)