Which are the solutions of the quadratic equation?

denis-greek [22]

Answer:

c. 441ft^2

Step-by-step explanation:

Area is height times length and since it's a square, safe to say that the sides are equal in length. So square root both of the given areas of the 2 square:

$\sqrt{841}$ = 29

$\sqrt{400}$ = 20

Now u have side b and c of the triangle, it's a right triangle so use the Pythagorean theorem a^2 + b^2 = c^2

a^2 + b^2 = c^2

a^2 + 20^2 = 29^2

a^2 + 400 = 841

subtract 4 from both sides

a^2 = 441

square root both sides

$\sqrt{a^2} = \sqrt{441}$

a = 21

Now the side a of the triangle is also a side of the small square therefore

21^2 = 441

8 0

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 $x-13$ .