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2 years ago

Question 8 of 10

Mathematics

1 answer:

Ymorist [56]2 years ago

5 0

Answer:

30 degrees

Step-by-step explanation:

100+50=150

180-150=30

All triangles equal 180 degrees, so if the other angles equal 150 than the third angle equals 30 degrees.

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Help please! (See image.)

Scorpion4ik [409]

FED is 5/4 times bigger than CBA. So, 44×1.25 is 55. 40×1.25 is 50. The perimeter is 125

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2 years ago

2x + 3y = 10

joja [24]

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2 years ago

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Jimmy applied the distributive property to write the equation below. 24 + 6 = 6 (4 + 2)

joja [24]

Answer: if you simplify this it becomes 30=28 but this is a false statement so you would say 30≠28

Step-by-step explanation:

≠ means not equal

7 0

1 year ago

Write each rational number as a terminating decimal. 1. 19/20 2. -1/8 3. 17/5

Dima020 [189]

19/20= 0.95
-1/8= -0.125
17/5= 3.4
HOPE THIS HELPS!

8 0

3 years ago

Read 2 more answers

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