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3 years ago

Can someone help me find the area and perimeter please?

Mathematics

1 answer:

Lelu [443]3 years ago

7 0

Answer:

Step-by-step explanation:

Remember that the formula to get the area of a right triangle is base times height divided by 2, so to find the area:

5 x 12 = 60/2 = 30

Area of this triangle is 30.

To find the perimeter, add all the sides:

5 + 12 + 13 = 30.

Perimeter of this triangle is 30.

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Help fast 70 points will give brainly if you give right answers

lisabon 2012 [21]

Answer:

answer below.

Step-by-step explanation:

hey, it's a little blurry so i hope i got these correct:)

1. rational

2. natural, whole, integer, rational

3. irrational

4. integer, rational

5. rational

6. integer, rational

7. 2

8. - π

9. -2

10. 1

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6 0

3 years ago

PLEASE HELP ME! the original problem is: p^2+4p-27 ÷ p+8

klasskru [66]

It is 6.75 i hope this helps i am NOT a 100% sure

3 0

3 years ago

The length of a rectangle is 4n centimeters, and it is twice as long as the width. The perimeter of the rectangle is twice as lo

sladkih [1.3K]

4n - the length of a rectangle
4n : 2 = 2n - the width of a rectangle

P = (4n + 2n) * 2
P = 12n

12n : 2 = 6n - the perimeter of the equilateral triangle
P = 3a
6n = 3a
a = 6n : 3
a = 2n - the side length of the triangle - answer.

6 0

4 years ago

Find the equation of the line that passes through (0, -3) and is parallel to

Tresset [83]

Hey there!

$\\$

Answer:

$\green{\boxed{\red{\bold{\sf{y = \dfrac{7}{6}x - 3}}}}}$

$\\$

Explanation:

To find the equation of a line, we first have to determine its slope knowing that parallel lines have the same slope.

Let the line that we are trying to determine its equation be $\: \sf{d_1} \:$ and the line that is parallel to $\: \sf{d_1} \:$ be $\: \sf{d_2} \:$ .

$\sf{d_2} \:$ passes through the points (9 , 2) and (3 , -5) which means that we can find its slope using the slope formula:

$\sf{m = \dfrac{\Delta y}{\Delta x} = \dfrac{\green{y_2} - \orange{y_1}}{\red{x_2} - \blue{x_1 }}}$

$\\$

⇒Subtitute the values :

$\sf{(\overbrace{\blue{9}}^{\blue{x_1}}\: , \: \overbrace{\orange{2}}^{\orange{y_1}}) \: \: and \: \: (\overbrace{\red{3}}^{\red{x_2}} \: , \: \overbrace{\green{-5}}^{\green{y_2}} )}$

$\implies \sf{m = \dfrac{\Delta y}{\Delta x} = \dfrac{\green{-5} - \orange{2}}{\red{ \: \: 3} - \blue{9 }} = \dfrac{ - 7}{ - 6} = \boxed{ \bold{\dfrac{7}{6} }}}$

$\sf{\bold{The \: slope \: of \: both \: lines \: is \: \dfrac{7}{6}}}$ .

Assuming that we want to get the equation in Slope-Intercept Form, let's substitute m = 7/6:

Slope-Intercept Form:

$\sf{y = mx + b} \\ \sf{Where \: m \: is \: the \: slope \: of \: the \: line \: and \: b \: is \: the \: y-intercept.}$

$\implies \sf{y = \bold{\dfrac{7}{6}}x + b}$ $\\$

We know that the coordinates of the point (0 , -3) verify the equation since it is on the line $\: \sf{d_1} \:$ . Now, replace y with -3 and x with 0:

$\implies \sf{\overbrace{-3}^{y} = \dfrac{7}{8} \times \overbrace{0}^{x} + b} \\ \\ \implies \sf{-3 = 0 + b} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \sf{\boxed{\bold{b = -3}} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$