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3 years ago

5

1.7 degrees Fahrenheit drops every thousand feet. The temperature on the ground is 63.7 Fahrenheit. What will the temperature be

when the plane reaches an altitude of 2,000 ft?

1 answer:

Oksanka [162]3 years ago

8 0

Answer:

The answer to your question is Final temperature = 60.3°F

Step-by-step explanation:

I include the process in the attachment

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(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

Montano1993 [528]

Answer:

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

Step-by-step explanation:

Given:

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

1.

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

2.

$q^0=1$

3.

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

4.

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

5.

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

6.

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

7.

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

8.

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

8 0

3 years ago

What is the correct answer of 1/3

serious [3.7K]

The correct answer of 1/3 is the decimal form of it which is 0.3 repeated (you can put a line on top of the 3). If not just leave it as 1/3 it is simplified as a fraction.

8 0

3 years ago

3. If the legs of an isosceles right triangle are 10 inches long, approximate the length of

Pavel [41]

Answer:

Using the Pythagorean theorem for isosceles right triangle, the length of the hypotenuse is sqrt( 2 x 10^2) = ~14.1 in

Hope this helps!

:)

6 0

3 years ago

Solve by using square root: <br> 4(x-1)^2+2=10

Scilla [17]

$4(x-1)^2+2=10\\\\4(x^2-2x+1)+2-10=0\\\\4 x^2-8x+4+2-10=0\\\\4 x^2-8x-4=0\ \ / :4\\\\x^2-2x-1=0\\\\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{2-\sqrt{ (-2)^2-4*1*(-1)}}{2 }=\frac{2-\sqrt{ 8}}{2 }=\frac{2- \sqrt{ 4*2}}{2 }=\\\\=\frac{2-\sqrt{ 8}}{2 }=\frac{2(1- \sqrt{ 2})}{2 }=1- \sqrt{ 2}\\\\x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{2+\sqrt{ (-2)^2-4*1*(-1)}}{2 }=\frac{2(1+ \sqrt{ 2})}{2 }=1+ \sqrt{ 2}$

$Answer:\ x=1- \sqrt{2}\ \ or\ \ x= 1+ \sqrt{ 2}$

3 0

4 years ago

Read 2 more answers

Please answer it as fast as u can

Vadim26 [7]

Answer:

Am I missing something to this question?

As pictured, the answer is 12 7/9ths.

7 0

3 years ago