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2 years ago

Triangle GHK has an area of 117cm2. Write and solve an equation to find the height of the triangle GHK

Mathematics

1 answer:

stellarik [79]2 years ago

6 0

Answer:

h=58.5/b

Step-by-step explanation:

This isn't necessarily possible, unless you know the base, but I can solve it with what we have, understanding the answer will contain variables:

The formula for the area of a triangle is A=1/2bh

117=(1/2)bh

58.5=bh

h=58.5/b

I'm not sure if this is the answer you were looking for, but I hope it helps

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Answer:

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Read 2 more answers

Find the area and perimeter of the square 6 meters

Mkey [24]

The formula for finding area of a square is:

$length \times width$

In a square, all the sides are the same. Each side is 6 meters long. Plug in 6 to the formula.

$length \times width \rightarrow 6 \times 6$

Multiply:

$6 \times 6 = 36$

The area of the square is 36 m²

<h2>PERIMETER</h2>

Perimeter is the distance around a shape. To find perimeter, you have to add all all the side lengths. The formula for finding perimeter is:

$2(l) \times 2(w)$

Remember that squares have 4 sides, and all length(s) and width(s) measure the same. The length and width measure the same. Plug in the numbers into the formula:

$2(l) + 2(w) \rightarrow 2(6) + 2(6)$

Simplify:

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Add:

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The perimeter of the square is 24 m

7 0

3 years ago

6-10 divide, another onee thank you!!

eduard

Answers:

10.) $\displaystyle \pm{5}$

9.) $\displaystyle 1\frac{1}{2}$

8.) $\displaystyle \pm{1\frac{1}{2}}$

7.) $\displaystyle \pm{1\frac{1}{2}}$

6.) $\displaystyle \pm{\frac{1}{2}}$

Step-by-step explanations:

10.) $\displaystyle \frac{\sqrt{200}}{\sqrt{8}} \hookrightarrow \sqrt{25} \hookrightarrow \frac{\pm{10\sqrt{2}}}{\pm{2\sqrt{2}}} \\ \\ \boxed{\pm{5}}$

9.) $\displaystyle \frac{\sqrt[3]{135}}{\sqrt[3]{40}} \hookrightarrow \sqrt[3]{3\frac{3}{8}} \hookrightarrow \frac{3\sqrt[3]{5}}{2\sqrt[3]{5}} \\ \\ \boxed{1\frac{1}{2}}$

8.) $\displaystyle \frac{\sqrt[4]{162}}{\sqrt[4]{32}} \hookrightarrow \sqrt[4]{5\frac{1}{16}} \hookrightarrow \frac{\pm{3\sqrt[4]{2}}}{\pm{2\sqrt[4]{2}}} \\ \\ \boxed{\pm{1\frac{1}{2}}}$

7.) $\displaystyle \frac{\sqrt{63}}{\sqrt{28}} \hookrightarrow \sqrt{2\frac{1}{4}} \hookrightarrow \frac{\pm{3\sqrt{7}}}{\pm{2\sqrt{7}}} \\ \\ \boxed{\pm{1\frac{1}{2}}}$

6.) $\displaystyle \frac{\sqrt{12}}{\sqrt{48}} \hookrightarrow \sqrt{\frac{1}{4}} \hookrightarrow \frac{\pm{2\sqrt{3}}}{\pm{4\sqrt{3}}} \\ \\ \boxed{\pm{\frac{1}{2}}}$

I am joyous to assist you at any time.

5 0

2 years ago