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3 years ago

It's possible to build a triangle with side lengths of 4, 4, and 16

Mathematics

2 answers:

lubasha [3.4K]3 years ago

5 0

Yes. It is possible

Stells [14]3 years ago

4 0

Answer:

Yes it's possible to make a triangle with 4, 4 and 16.

But if it also has 2 equal angles along with the equal sides (4 and 4), you can call it isosceles triangle.

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Calculate the size of each interior angle of a regular polygon with 14 sides

Ierofanga [76]

Answer:

154.285

Step-by-step explanation:

The sum of the interior angles of a 14-gon = (14-2)*180 = 2,160 degrees. There are 14 vertexes (vertices) for a 14-gon. It is 'regular' so all these angles are equal. By dividing 2160 (sum of all angles) by 14 (total amount of angles) we can see how much each angle is worth. So the interior angle of each is 154.285 degrees.

4 0

3 years ago

What is the answer to the pic

MA_775_DIABLO [31]

Answer:

$7$

Step-by-step explanation:

$\frac{7^{-1}}{7^{-2}}$

$7^{-1-(-2)}$

$7^{-1+2}$

$7^{1}$

$7$

I used the following rules:

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

$a^1=a$

$a \neq 0$

6 0

3 years ago

darren jogs 425 meters and runs 650 meters in one workout.he performs this workout three times .what is the total distance cover

Igoryamba

425+650=1075 1075x3=3225

4 0

3 years ago

Read 2 more answers

The bar diagram represents 24 hours in one day and shows a mark at 30%If you sleep 30% of the day, how many hours do you sleep?

horrorfan [7]

Let:

100%---------------->24h

30%------------------>xh

Using cross multiplication:

$\begin{gathered} \frac{100}{30}=\frac{24}{x} \\ solve_{\text{ }}for_{\text{ }}x\colon \\ x=\frac{30\cdot24}{100} \\ x=\frac{720}{100}=7.2h \end{gathered}$

8 0

1 year ago

You are in charge of managing the advertising for a restaurant. You are able to buy up to 62 advertisements a month, and your da

erik [133]

Answer:

Greatest number of newspaper ads = 59

Greatest number of television ads = 23

Step-by-step explanation:

It is given that we should buy at least 5 newspapers for every 3 television ads.

So, after purchasing 3 television ads, we can buy any number of television ads but the number should be greater than 5.

Since, the maximum is 62, we can buy 59 newspaper ads. (Note that we have already bought 3 television ads).

For every 3 television ads, we should buy 5 newspaper ads. So,

For the purchase of 3 television ads, we should buy the minimum of 5 newspaper ads.

For the purchase of 2(3) television ads, we should buy the minimum of 2(5) newspaper ads.

For the purchase of 3(3) television ads, we should buy the minimum of 3(5) newspaper ads.

...

For the purchase of 7(3) = 21 television ads, we should buy the minimum of 7(5) = 35 newspaper ads.

So, totally 21+35 = 56 ads.

For the remaining 6 ads, if we buy 3 television ads, we should buy 5 newspaper ads.

So, we can buy 2 television ads and 4 newspaper ads.

Hence, we can buy the maximum of 21+2 = 23 television ads.

7 0

3 years ago

Read 2 more answers