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3 years ago

Which graph best shows the relationship between the length of one side of a square and its perimeter.

Mathematics

2 answers:

sladkih [1.3K]3 years ago

6 0

The answer is the 1st graph. This is because a square have 4 sides and it is all the same. on the x axis (Side length) when it is 4 the perimeter (Length around the outside) will be 4 x 4 which is 16.

Hope this helps?

Anna [14]3 years ago

5 0

Answer:

<h2>First graph.</h2>

Step-by-step explanation:

Remember that a square has all four equal sides, which we are gonna call $x$ .

Now, the perimeter is defined as the sum of all sides, which is $P=x+x+x+x=4x$

As you can notice, the perimeter of a square is four times its side. If we call $y$ the perimeter, the relation is

$y=4x$

That means if each side is 4 units, then its perimeter is 16 units.

If you observe each graph, you would find that the first graph describes this relation. To demonstrate that, let's the slope of the line using (0,0) and (4, 16) which are two points on the line

$m=\frac{16-0}{4-0}=\frac{16}{4}=4$

Notice that the first graph has the same slope (ratio of change) than the expression we deducted.

Therefore, the right answer is the first graph.

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Answer:

Step-by-step explanation:

First, let's work out the two triangles:

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Area of rectangle: l x w

3 x 8 = 24

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Answer:

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

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3 years ago

Find the square roots of each complex number. a. 5 + 12i b. 5 − 12i

musickatia [10]

ANSWER:

Square root of 5 + 12i is 3+2i or -3 – 2i and square root of 5 – 12i is 3 – 2i or -3 + 2i.

SOLUTION:

Given, we have to find square roots of 5 + 12i and 5 – 12i

a) 5 + 12i

Now, square root of 5 + 12i

Suppose that a + bi is a square root of 5 + 12i.

$\text { Then, }(a+b i)^{2}=\left(a^{2}-b^{2}\right)+(2 a b) i=5+12 i$

Equate real and imaginary parts:

$\begin{array}{l}{a^{2}-b^{2}=5 \text { and } 2 a b=12 \rightarrow b=\frac{6}{a}} \\\\ {\text { So, } a^{2}-\left(\frac{6}{a}\right)^{2}=5} \\\\ {\rightarrow a^{2}-\frac{36}{a^{2}}=5} \\\\ {\rightarrow a^{4}-5 a^{2}-36=0} \\\\ {\rightarrow\left(a^{2}-9\right)\left(a^{2}+4\right)=0}\end{array}$

Since a must be real, a = 3 or -3.

This gives b = 2 or -2, respectively.

Thus, we have two square roots: 3+2i or -3-2i.

b) 5 - 12i

Now, square root of 5 - 12i

Suppose that a - bi is a square root of 5 - 12i.

$\text { Then, }(a-b i)^{2}=\left(a^{2}-b^{2}\right)-(2 a b) i=5-12 i$

Equate real and imaginary parts:

$\begin{array}{l}{a^{2}-b^{2}=5 \text { and } 2 a b=12 \rightarrow b=\frac{6}{a}} \\\\ {S o, a^{2}-\left(\frac{6}{a}\right)^{2}=5} \\\\ {\rightarrow a^{2}-\frac{36}{a^{2}}=5} \\\\ {\rightarrow a^{4}-5 a^{2}-36=0} \\\\ {\rightarrow\left(a^{2}-9\right)\left(a^{2}+4\right)=0} \\\\ {\text { since a must be real, } a=3 \text { or }-3}\end{array}$

This gives b = 2 or -2, respectively.

Thus, we have two square roots: 3 - 2i or -3 + 2i.

Hence, square root of 5 + 12i is 3+2i or -3 – 2i and square root of 5 – 12i is 3 – 2i or -3 + 2i.

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3 years ago