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3 years ago

Solve for k : 62=13k-3

1 answer:

4 0

$62=13k-3$

$62+3=13k

$ $\text{Add 3 to both sides}$

$65=13k
$

$65 \div 13 = k$ $\text{Divide both sides by 13}$

$k=5$ $\text{Solution}$

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An isosceles triangle has an angle that measures 100°. Which other angles could be in that isosceles triangle?

HACTEHA [7]

Answer:

40°

Step-by-step explanation:

because all triangles and up to 180°

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3 years ago

In the cordinate plane what is the length of the line segment that connects points at (0, -1) and (-7, -2) ? Enter your answer i

allsm [11]

Answer:

≈ 7.07

Step-by-step explanation:

Calculate the length using the distance formula

d = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (0, - 1) and (x₂, y₂ ) = (- 7, - 2)

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4 years ago

-15w -6w +7w=14 solve for w

bearhunter [10]

Answer:

w=-1

Step-by-step explanation:

-15w-6w+7w=14

-21w+7w=14

-14w=14

w=14/-14

w=-1

7 0

3 years ago

Read 2 more answers

A Norman window is a window with a semi-circle on top of regular rectangular window. (See the picture.) What should be the dimen

Vikki [24]

Answer:

bottom side (a) = 3.36 ft

lateral side (b) = 4.68 ft

Step-by-step explanation:

We have to maximize the area of the window, subject to a constraint in the perimeter of the window.

If we defined a as the bottom side, and b as the lateral side, we have the area defined as:

$A=A_r+A_c/2=a\cdot b+\dfrac{\pi r^2}{2}=ab+\dfrac{\pi}{2}\left (\dfrac{a}{2}\right)^2=ab+\dfrac{\pi a^2}{8}$

The restriction is that the perimeter have to be 12 ft at most:

$P=(a+2b)+\dfrac{\pi a}{2}=2b+a+(\dfrac{\pi}{2}) a=2b+(1+\dfrac{\pi}{2})a=12$

We can express b in function of a as:

$2b+(1+\dfrac{\pi}{2})a=12\\\\\\2b=12-(1+\dfrac{\pi}{2})a\\\\\\b=6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a$

Then, the area become:

$A=ab+\dfrac{\pi a^2}{8}=a(6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a)+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a^2+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}-\dfrac{\pi}{8}\right)a^2\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{8}\right)a^2$