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3 years ago

I need help with this please

Mathematics

1 answer:

Shalnov [3]3 years ago

7 0

I think that the answer is 10 sorry if l be wrong

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Which is an estimate of the square root of 14 to the nearest hundredth?

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3 years ago

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How do I solve and whats the answer?

8090 [49]

Answer:

x = -6

Step-by-step explanation:

Here we have an equilateral triangle. All three sides have the same length, 2. Therefore, the interior angles are all equal to 60 degrees.

Therefore, 6 - 9x = 60, and -9x = 54. Dividing both sides by -9 results in

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3 years ago

(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

Montano1993 [528]

Answer:

$\dfrac{m^3}{16p^2q^2}$

Step-by-step explanation:

Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

$m^{-1}=\dfrac{1}{m}$

$q^0=1$

$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

$(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}$

$m^{-1}=\dfrac{1}{m}$

$2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}$

$\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}$

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}$

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3 years ago

Number 1 les goooooooo

chubhunter [2.5K]

Answer:

Hi thank you so muchhhhh

Step-by-step explanation:

I wish you a great day

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2 years ago

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Write 17000 in decreasing order

Brrunno [24]

Is it 00017?????? Just guessing

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3 years ago

I need help with this please​

I need help with this please