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2 years ago

Find Perimeter: SQUARE with sides that are each 5 inches in length

Mathematics

2 answers:

balandron [24]2 years ago

7 0

Answer:

20 inches

Step-by-step explanation:

a square has 4 equal sides, so you can add 5+5+5+5 to find the sum of all the side lengths.

seraphim [82]2 years ago

6 0

If we’re talking 2d it would be 20inches, 3D it would be 120inches

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juin [17]

Answer:

five , seven

Step-by-step explanation:

brainlyist

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2 years ago

3 3/5 + 1 1/4 = ??????

Nikitich [7]

Answer:

4 & 17/20

Step-by-step explanation:

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3 years ago

Given the formula a = s2 + 3b - c , find the value of a when s = 7, b = 13 and c = -24.

galben [10]

Answer:

a = 112

Step-by-step explanation:

$a =s^{2} +3b-c$

Given that,

s = 7

b = 13

c = (-24)

Let's solve for a now.

$a =s^{2} +3b-c$

$a = 7^{2} +3*13-(-24)\\$

$a=49+39+24\\$

$a=88+24$

$a=112$

Hope this helps you.

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3 0

2 years ago

In a class of 34 students, 19 of them are girls.

Colt1911 [192]

Answer:

55.9%

Step-by-step explanation:

To find the percent that are girls, take the number of girls over the total number of students

19/34

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To change to percent form, multiply by 100%

55.8823529%

To 1 decimal place

55.9%

5 0

3 years ago

Read 2 more answers

Solve the following equation by factoring:9x^2-3x-2=0

olya-2409 [2.1K]

Answer:

The two roots of the quadratic equation are

$x_1= - \frac{1}{3} \text{ and } x_2= \frac{2}{3}$

Step-by-step explanation:

Original quadratic equation is $9x^{2}-3x-2=0$

Divide both sides by 9:

$x^{2} - \frac{x}{3} - \frac{2}{9}=0$

Add $\frac{2}{9}$ to both sides to get rid of the constant on the LHS

$x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}$ ==> $x^{2} - \frac{x}{3}=\frac{2}{9}$

Add $\frac{1}{36}$ to both sides

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{2}{9} +\frac{1}{36}$

This simplifies to

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}$

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = $\frac{1}{6}\right)$ we can see that

$\left(x - \frac{1}{6}\right)^2$ = $x^2 - 2.x. (-\frac{1}{6}) + \frac{1}{36} = x^{2} - \frac{x}{3}+\frac{1}{36}$

$\left(x - \frac{1}{6}\right)^2=\frac{1}{4}$

Taking square roots on both sides

$\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}$

So the two roots or solutions of the equation are

$x - \frac{1}{6}=-\sqrt{\frac{1}{4}}$ and $x - \frac{1}{6}=\sqrt{\frac{1}{4}}$

$\sqrt{\frac{1}{4}} = \frac{1}{2}$

So the two roots are

$x_1=\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}$

and

$x_2=\frac{1}{6} + \frac{1}{2} = \frac{2}{3}$

7 0

1 year ago