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tankabanditka [31]

3 years ago

15

Determine whether the measures 9, 40, and 41 can be the measures of the sides of a triangle, If so, classify the triangle as ocu

te, obtuse, or right.
Justify your answer.

1 answer:

melamori03 [73]3 years ago

7 0

Answer:

B) yes, right

Step-by-step explanation:

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What is the discount price os a skateboard that cost $155.80 on sale for 20% off

pshichka [43]

Here's an example of %
20% = 0.2
2% = 0.02

when multiplying with decimals in place you always add up the two periods, 155.8 × 0.2 = 31.16
since the dot is one over for both you add 1 + 1 which equals 2, then when you get 31.16 you move over 2 decimal points, making it 31.16
btw the answer is when you subtract 31.16 from 155.80 which would be 124.64$ for your final cost.

6 0

3 years ago

What is 20 times 98 is it a b c d

Lyrx [107]

20 times 98 is the product of 20 and 98

Thus 20 times 98 = 20 × 98 = 1960

5 0

4 years ago

-2c + 3c - 5 - 4c + 7

kaheart [24]

Answer:−

3c+2

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

3/4n=675 what is the answer to N

densk [106]

1) 506.25
2) 20
3) x = -1
4) x = 5
Hope this Helps

6 0

3 years ago

Explain me PLEASEEEE!!!

kirza4 [7]

Answer:

$\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10}}=\frac{a^{3}}{a^{10}}=a^{-7}=\frac{1}{a^{7}}$

Step-by-step explanation:

Let us revise the properties of exponents

$a^{m}.a^{n}=a^{m+n}$

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

$(a^{m})^{n}=a^{m.n}$

$a^{-m}=\frac{1}{a^{m} }$

Let us use these properties to solve the question

→ By using the 3rd property above

∵ $(a^{2})^{3}=a^{2.3}=a^{6}$

∴ $\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10} }$

→ By using the 1st property above

∵ $a^{6}.a^{-3}=a^{6+-3}=a^{6-3}=a^{3}$

∴ $\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10}}=\frac{a^{3}}{a^{10}}$

→ By using the 2nd property above

∵ $\frac{a^{3}}{a^{10}}=a^{3-10}=a^{-7}$

∴ $\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10}}=\frac{a^{3}}{a^{10}}=a^{-7}$

→ By using the 4th property above

∵ $a^{-7}=\frac{1}{a^{7}}$

∴ $\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10}}=\frac{a^{3}}{a^{10}}=a^{-7}=\frac{1}{a^{7}}$

8 0

3 years ago