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3 years ago

What is the area of a regular nonagon with 10 cm side length and an apothem of 6 cm?

Mathematics

1 answer:

pickupchik [31]3 years ago

8 0

The area of the regular nonagon is 270 sq cm.

Step-by-step explanation:

Given,

Each side of a regular nonagon (b) = 10 cm

The length of apothem (h) = 6 cm.

To find the area of the nonagon.

Formula

The area of a nonagon with b as each side and h as apothem is = 9( $\frac{1}{2}$ bh)

Now,

Putting the value of h and b we get,

Area = 9( $\frac{1}{2}$ ×10×6) sq cm = 270 sq cm

Hence, the area is 270 sq cm.

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Daisy spent a total of $31 before tax on ice cream. the table below shows the different flavors she bought. a carton of strawber

notsponge [240]

The cost of 1 vanilla ice cream and 1 strawberry ice cream should total to $31. Since you only bought one of each flavor, you simply have to add their unit prices.

Cost for 1 carton of strawberry ice cream + Cost for 1 carton of vanilla ice cream = Total Cost
Total Cost = s + v --> this is the general equation

Since s = 1.5 + v, the modified equation would be
Total Cost = 1.5 + v + v
Total Cost = 1.5 + 2v

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3 years ago

a recipe for bread uses 5 3/4 cups of white flour and 3 1/3 cups of wheat flour. How many more cups of white flour than wheat fl

zepelin [54]

2 and 3/4 more cups of white flour are used than wheat flour.
5 x 3/4= 3 3/4
3 x 1/3= 1
3 3/4 - 1+ 2 3/4

4 0

3 years ago

Read 2 more answers

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

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3 years ago

Giving brainless! Don’t answer if you don’t know I’ll flag you.

GenaCL600 [577]

Answer:

The sales tax is $0.40

Step-by-step explanation:

10 x 0.04 = 0.4

0.04 is 40% when multiplying

4 0

3 years ago

How are 2-D and 3-D objects related to each other in real world

stich3 [128]

Answer:

You can see 2-D flat shapes on the faces of 3-D objects.

For example: You go to the store and you see a box,

the box has 6 faces. The bottom of the box is one of It's faces which is 2-D.

Step-by-step explanation:

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3 years ago