All categories

3 years ago

If a polygon is a square or a rectangle one may find the perimeter by

Mathematics

1 answer:

DaniilM [7]3 years ago

5 0

Périmètre rectangle = 2 ( longueur + largeur)

périmètre carré = 4 x côté

You might be interested in

Solve the simultaneous equations:<br> 3x + 2y = 35 <br> 2x + 3y = 30

lesya [120]

Answer:

nuber 1

Simplifying

3x + 2y = 35

Solving

3x + 2y = 35

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-2y' to each side of the equation.

3x + 2y + -2y = 35 + -2y

Combine like terms: 2y + -2y = 0

3x + 0 = 35 + -2y

3x = 35 + -2y

Divide each side by '3'.

x = 11.66666667 + -0.6666666667y

Simplifying

x = 11.66666667 + -0.6666666667y

5 0

3 years ago

I need help with number 20 please

Doss [256]

Answer:

3 = 92 degree 4 = 88 degree

Step-by-step explanation:

3 = 45 + 47 = 92 degree

4 = 180 - 92 = 88 degree

5 0

3 years ago

Read 2 more answers

ABC is isosceles.<br> m X = [?]<br> please help!!

Rashid [163]

Answer:

x = 4

Step-by-step explanation:

The base angles of an isosceles triangle are congruent, as shown in the figure.

Equating their measures, we have ...

∠A = ∠C

8x -7 = 25 . . . . substitute the given expressions

8x = 32 . . . . . add 7

x = 4 . . . . . . divide by 8

6 0

2 years ago

Stacy and Neil painted 280 ft2 of their living room wall with 4/5 gallon of paint. How many square feet can they paint with 4 ga

Neporo4naja [7]

Answer:

Step-by-step explanation:

Divide 280 by 4/5 and get 14 then multiply by 4 and get 56

3 0

3 years ago

Read 2 more answers

If cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2π, what are the values of sin Θ and tan Θ?

KIM [24]

Answer:

The answer is

$sin(\theta)=-\frac{\sqrt{2}}{2}$

$tan(\theta)=-1$

Step-by-step explanation:

we know that

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

$sin^{2}(\theta)+cos^{2}(\theta)=1$

In this problem we have

$cos(\theta)=\frac{\sqrt{2}}{2}$

$\frac{3\pi}{2}$

The angle $\theta$ belong to the third or fourth quadrant

The value of $sin(\theta)$ is negative

Step 1

Find the value of $sin(\theta)$

Remember

$sin^{2}(\theta)+cos^{2}(\theta)=1$

we have

$cos(\theta)=\frac{\sqrt{2}}{2}$

substitute

$sin^{2}(\theta)+(\frac{\sqrt{2}}{2})^{2}=1$

$sin^{2}(\theta)=1-\frac{1}{2}$

$sin^{2}(\theta)=\frac{1}{2}$

$sin(\theta)=-\frac{\sqrt{2}}{2}$ ------> remember that the value is negative

Step 2

Find the value of $tan(\theta)$

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

we have

$sin(\theta)=-\frac{\sqrt{2}}{2}$

$cos(\theta)=\frac{\sqrt{2}}{2}$

substitute

$tan(\theta)=\frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}$

$tan(\theta)=-1$

8 0

3 years ago

Read 2 more answers