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3 years ago

Solve the problem below and enter your answer in the space below 7x (3-2) = ?

Mathematics

1 answer:

postnew [5]3 years ago

5 0

Answer:

Step-by-step explanation:

PEMDAS

3-2=1

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A map is drawn so that every 1 inch on the map represents 50 actual miles. The table below can be used to record the actual dist

olasank [31]

200 miles
You just multiply 50 by 4.

:) have a good day !

7 0

3 years ago

Jaxson buys ice cream and apples at the store. He pays a total of $24.81. He pays a total of $3.60 for the ice cream. He buys 7

Vadim26 [7]

Answer:

$3.03

Step-by-step explanation:

7x + $3.60 = $24.81

Subtract $3.60 from both sides of the equation:

7x + $3.60 - $3.60 = $24.81 - $3.60

7x = $24.81 - $3.60

7x = $21.21

Divide each side by 7:

(7x/7) = ($21.21/7)

x = ($21.21/7)

x = $3.03

3 0

3 years ago

D=ABC for A solve for A

Dafna1 [17]

A = D/BC. In order to isolate A, all you do is divide by BC

7 0

3 years ago

Differentiate with respect to X <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7Bcos2x%7D%7B1%20%2Bsin2x%20%7D%20

Mice21 [21]

Power and chain rule (where the power rule kicks in because $\sqrt x=x^{1/2}$ ):

$\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'$

Simplify the leading term as

$\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}$

Quotient rule:

$\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'=\dfrac{(1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))'}{(1+\sin(2x))^2}$

Chain rule:

$(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)$

$(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)$

Put everything together and simplify:

$\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}$

$=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x)}{(1+\sin(2x))^2}$

$=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2}{(1+\sin(2x))^2}$

$=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac{\sin(2x)+1}{(1+\sin(2x))^2}$

$=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}$

$=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}$

$=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}$

5 0

3 years ago

If the radius of the cylinder is doubled does the volume of the cylinder double?

Anastaziya [24]

The answer to your question is yes

6 0

3 years ago