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3 years ago

Umm yah i know I suck at math

Mathematics

1 answer:

nasty-shy [4]3 years ago

4 0

Its D because look if u add the white and red ones its 14 and all them together is 26 so it would be 14/26

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Find each product. (2.05).(0.004)

AlladinOne [14]

Answer:

$(2.05).(0.004) \\ = ( \frac{205}{100} ).( \frac{4}{1000} ) \\ = \frac{820}{100000} \\ = \frac{82}{10000} \\ = 0.0082$

3 0

3 years ago

Simplify 3(2x - 5). -6x - 15 6x - 15 6x - 8 5x - 2

Alekssandra [29.7K]

You just times it. (if you have a more question pm me). thank you

5 0

2 years ago

Read 2 more answers

Which costs more per gallon? 5 gallons for $12.50 or 11 gallons for $26.40

Effectus [21]

Answer: 5

Step-by-step explanation:

If;

5 gallons = $12.50 / 5 = $2.50/gallon

11 gallons = $26.40 / 11 = $2.40/gallon

In conclusion, 5 gallons cost more per gallon than 11.

6 0

3 years ago

Differentiate with respect to X <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

2 years ago

7 times 29 using area models

drek231 [11]

Answer:

949

Step-by-step explanation:

8 0

2 years ago