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4 years ago

What is the product (4s+2)(5s^2+10s+3)

Mathematics

2 answers:

Greeley [361]4 years ago

5 0

Answer:

$20s^3+50s^2+32s+6$

Explanation:

we have been given the function $\left(4s+2\right)\left(5s^2+10s+3\right)$

So, as to multiply the two expression step by step calculation is shown below:

First of all we will open the parenthesis

$4s\cdot \:5s^2+4s\cdot \:10s+4s\cdot \:3+2\cdot \:5s^2+2\cdot \:10s+2\cdot \:3$

$4\cdot \:5s^2s+4\cdot \:10ss+4\cdot \:3s+2\cdot \:5s^2+2\cdot \:10s+2\cdot \:3$

After simplification we will get the following:

$4\cdot \:5s^2s+4\cdot \:10ss+4\cdot \:3s+2\cdot \:5s^2+2\cdot \:10s+2\cdot \:3:\quad 20s^3+50s^2+32s+6$

After further simplification we will get

$20s^3+50s^2+32s+6$

neonofarm [45]4 years ago

3 0

20s^3 + 40s^2 + 12s + 10s^2 + 20s + 6
20s^3 + 50s^2 + 32s + 6

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Which shows 60^2 − 40^2 being evaluated using the difference of squares method?

DerKrebs [107]

D. The answer has to be 2,000, eliminating A and C. B doesn't even use the original equation, as well as incorrectly solving the method, so D is the correct answer, even if you don't understand the principles.

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romanna [79]

Answer:

Step-by-step explanation:

MR=2x-1. First equation

RP=9x+3 second equation

MP=MR+RP third equation

MP= 2x-1+9x+3 first equation plus second equation

MP=11x+2

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4 years ago

Trying to figure this out

prisoha [69]

I hope this helps you

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3 years ago

1. Find four consecutive even integers such that the sum of the first and third is 6 less than the largest

Sophie [7]

These are indeed quite a lot of exercises, but a lot of them are almost identical - only small computations will change. So, I'm glad to help you with one exercise from every cathegory, but I encourage you to solve the others on your own.

<h2>Exercise 1</h2>

You can call four consecutive integers as

$x,\ x+1,\ x+2,\ x+3$

So, the sum of the first and third is $x+(x+2) = 2x+2$

We want this quantity to be six less than the largest, i.e. $(x+3)-6 = x-3$

So, the equality is

$2x+2 = x-3$

Subtract x from both sides:

$x+2 = -3$

Subtract 2 from both sides:

$x = -5$

So, the consecutive integers are

$-5,\ -4,\ -3,\ -2$

In fact, the sum of the first and third is $-5-3 = -8$ , which is indeed six less than the largest: $-8 = -2-6$

<h2>Exercise 2</h2>

If you call the first odd number $x$ , the next consecutive odd numbers will be $x,\ x+2,\ x+4,\ x+6$

In fact, we have to count skipping two's, because we only want odd integers. From here, you go on like exercise 1: you write the largest (which is x+6), and set it to be two more than the sum of the other three (x, x+2 and x+4)

<h2>Exercise 3</h2>

By the same logic of exercise 2, two consecutive even integers are $x,\ x+2$ , assuming that x is even.

So, you set the equation as usual: the smaller (which is x) is 26 less than three times the larger (which means 3(x+2)-26)

<h2>Exercise 4 to 8</h2>

These are all pretty identical to exercise 1: you start by listing three or four consecutive integers:

$x,\ x+1,\ x+2\quad\text{or}\quad x,\ x+1,\ x+2\ x+3$

and then you translate the request of each exercise accordingly. Remember that expressions like "three times the second number" means that you have to multiply: $3(x+1)$ , while expression like "six more than the first" or "thirteen less than the first" imply adding/subtracting: $x+6$ or $x-13$ .