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2 years ago

Verify the Identity

Mathematics

1 answer:

valina [46]2 years ago

5 0

Answer:

Cos x sec^2 x

Cos x (1 + cot x)

Cos x / sin x • 1

Step-by-step explanation:

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I am lost on what to do

Neko [114]

$\bf sin({{ \alpha}})sin({{ \beta}})=\cfrac{1}{2}[cos({{ \alpha}}-{{ \beta}})\quad -\quad cos({{ \alpha}}+{{ \beta}})]
\\\\\\
cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}\\\\
-----------------------------\\\\
\lim\limits_{x\to 0}\ \cfrac{sin(11x)}{cot(5x)}\\\\
-----------------------------\\\\
\cfrac{sin(11x)}{\frac{cos(5x)}{sin(5x)}}\implies \cfrac{sin(11x)}{1}\cdot \cfrac{sin(5x)}{cos(5x)}\implies \cfrac{sin(11x)sin(5x)}{cos(5x)}$

$\bf \cfrac{\frac{cos(11x-5x)-cos(11x+5x)}{2}}{cos(5x)}\implies \cfrac{\frac{cos(6x)-cos(16x)}{2}}{cos(5x)}
\\\\\\
\cfrac{cos(6x)-cos(16x)}{2}\cdot \cfrac{1}{cos(5x)}\implies \cfrac{cos(6x)-cos(16x)}{2cos(5x)}
\\\\\\
\lim\limits_{x\to 0}\ \cfrac{cos(6x)-cos(16x)}{2cos(5x)}\implies \cfrac{1-1}{2\cdot 1}\implies \cfrac{0}{2}\implies 0$

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3 years ago

What is the shortest possible distance

Licemer1 [7]

Answer:

I think it can be 3 . Hope it help you

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2 years ago

Find the product. <br> (2xyz) (-4x2yz)<br><br> PLEASE HELP!!! ASAP!!!

melamori03 [73]

Answer:

$-16x^{2} y^{2}z^{2}$

Step-by-step explanation:

Let's calculate the answer to the second parenthesis. Just multiply throughout to get -8xyz. Then, multiply the first parenthesis by the second. Multiply the constants (-8 and 2) first to get -16, multiply the x's together to get x^2, the y's together to get y^2, and the z's together to get z^2. Put it all together to get $-16x^{2} y^{2}z^{2}$ .

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3 years ago

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What are the zeros in the function

Schach [20]

$f(x)=x^3+x^2-6x\\\\text{The zeros:}\\\\f(x)=0\to x^3+x^2-6x=0\\\\x(x^2+x-6)=0\\\\x(x^2+3x-2x-6)=0\\\\x[x(x+3)-2(x+3)]=0\\\\x(x+3)(x-2)=0\iff x=0\ \vee\ x+3=0\ \vee\ x-2=0\\\\x=0\ \vee\ x=-3\ \vee\ x=2$

Answer: -3; 0 and 2

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3 years ago

Solve for q. –10 + 10q = 9q

Kitty [74]

Answer:

q= 10

Step-by-step explanation:

9q - 10q = -10

-q = -10

q= -10/-1 = 10

6 0

3 years ago

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