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sergey [27]
3 years ago
6

Help please I do not get this problem at all

Mathematics
1 answer:
lara31 [8.8K]3 years ago
4 0

Answer:

It should be 4. The ratio that AC is equal to DE at should be equal to the ratio of BE to BC. If you want an explanation I'd be happy to add one!

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The pressure, P, of a gas, varies inversely with its volume, V. Pressure is measured in units of Pa. Suppose that a particular a
k0ka [10]
<span>The pressure, P, of a gas, varies inversely with its volume, V
</span>
∴ PV = k    where k is constant

<span>The  pressure is 84 Pa at a volume of 36 L.
</span>
<span>∴ at P = 84  ⇒⇒⇒  V = 36
</span>
<span>∴ K = PV = 84 * 36 = 3,024  ⇒⇒(1)</span>

If the volume is expanded to 216 L, what will the new pressure be?
We need to find P  at  V = 216
From the equation (1) ⇒⇒⇒ PV = 3024
substitute with V = 216
∴ 216 P = 3024
∴ P = 3024/216 = 14 

The correct answer is option <span>B. 14 Pa</span>



4 0
4 years ago
Please help meee!! I don’t understand this
Alina [70]
13 I guess, or 12 i don't rlly know
6 0
3 years ago
Read 2 more answers
Lowering powers write in terms of first power of cosine. Cos^6
yaroslaw [1]

The main identity you need is the double angle one for cosine:

\cos^2x=\dfrac{1+\cos2x}2

We get

\cos^6x=(\cos^2x)^3=\left(\dfrac{1+\cos2x}2\right)^3=\dfrac{(1+\cos2x)^3}8

Expand the numerator to apply the identity again:

\cos^6x=\dfrac{1+3\cos2x+3\cos^22x+\cos^32x}8

\cos^6x=\dfrac{1+3\cos2x+3\left(\frac{1+\cos2(2x)}2\right)+\cos2x\left(\frac{1+\cos2(2x)}2\right)}8

\cos^6x=\dfrac{1+3\cos2x+\frac32+\frac32\cos4x+\frac12\cos2x(1+\cos4x)}8

\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{16}\cos2x\cos4x

Finally, make use of the product identity for cosine:

\cos2x\cos4x=\dfrac{\cos6x+\cos2x}2

so that ultimately,

\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos2x+\dfrac1{32}\cos6x

\cos^6x=\dfrac5{16}+\dfrac{15}{32}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos6x

4 0
4 years ago
A-1/6<br> B-2/9<br> C-3/8<br> D-4/7
mash [69]

Answer:

3/8 is terminating decimal number. It's decimal value is 0.375

7 0
3 years ago
Find the surface area of a box measuring 4 inches long 3 inches wide and 6 inches tall
Rufina [12.5K]

4*3

12

12*6

72

hope it help

3 0
3 years ago
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