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Marat540 [252]
3 years ago
7

Answer This And Show Work Please ,?

Mathematics
1 answer:
skad [1K]3 years ago
6 0

Answer:

-11s^{5}v^{4}

Step-by-step explanation:

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

So you can apply here:

\frac{-11v^{6}s^{5}}{v^{2}} = -11s^{5}v^{6-2}

So your left with

-11s^{5}v^{4}

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I need help on this
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The width and height of a television screen are 80 in and 45 in, respectively. Find the area of this television screen.
r-ruslan [8.4K]

Answer: C: 3600


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3 years ago
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Solve the following equation
NISA [10]

Answer:

n = -13

Step-by-step explanation:

38 = -2n + 12

<em>Subtracting</em><em> </em><em>1</em><em>2</em><em> </em><em>from</em><em> </em><em>both</em><em> </em><em>sides</em><em>:</em>

==> 38 - 12 = -2n + 12 - 12

==> 26 = -2n

<em>multiplying both sides by (-1)</em><em>:</em>

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<em>dividing</em><em> </em><em>both</em><em> </em><em>sides by 2:</em>

==> -13 = n

3 0
3 years ago
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85.87 J or heat energy is added to a 34.8 g mass of substance the temperature rises from 21.76°C
Andru [333]

Answer:

The required specific heat is 196.94 joule per kg per °C  

Step-by-step explanation:

Given as :

The heat generated = Q = 85.87 J

Mass of substance  (m)= 34.8 gram = 0.0348 kg

Change in temperature = T2 - T1 = 34.29°C - 21.76°C = 12.53°C

Let the specific heat = S

Now we know that

Heat = Mass × specific heat × change in temperature

Or, Q = msΔt

Or, 85.87 = (0.0348 kg ) × S × 12.53°C

Or , 85.87 = 0.4360 × S

Or, S = \frac{85.87}{0.4360}

∴ S = 196.94 joule per kg per °C

Hence the required specific heat is 196.94 joule per kg per °C   Answer

7 0
3 years ago
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