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3 years ago

14

Given that HOP is similar to TAG, which of the following statements must be true? Check all that apply. A. sin T = sin H B. cos

P + cos G = 2cos P C. tan G = tan H D. tan T + tan P = 2tan T

2 answers:

pentagon [3]3 years ago

5 0

<u>The correct answers (Apex) are:</u>

- cos P + cost G = 2cos P

- sin T = sin H

Tju [1.3M]3 years ago

3 0

Pls. see attachment.

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Find the slope of y = log ⁡ ( x^3 ) at x = 1.

romanna [79]

i'm pretty sure the slope is zero ...

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3 years ago

Find the 53rd term of the arithmetic sequence<br> 27,11, -5

vivado [14]

Answer:

The 53rd term of this arithmetic sequence is -805.

Step-by-step explanation:

The general rule of an arithmetic sequence is the following:

$a_{n+1} = a_{n} + d$

In which d is the common diference between each term, that is, $d = a_{3} - a_{2} = a_{2} - a_{1}$ .

To find the nth term of the sequence, this equation can be written as:

$a_{n} = a_{1} + (n-1)d$

27,11, -5

So $a_{1} = 27, a_{2} - a_{1} = 11 - 27 = -16[/tex[tex]a_{n} = a_{1} + (n-1)d$

$a_{53} = a_{1} + (52)d = 27 + 52*(-16) = -805$

The 53rd term of this arithmetic sequence is -805.

3 0

2 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

Please help me pleasse

nirvana33 [79]

Answer:

75 mph

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Given the diagram below if the measure of <1 is 36° what is the

mina [271]

Answer:

144

Step-by-step explanation:

Because the horizontal lines are parallel <7 is the same as <4. Because <1 and <7 are on a straight line they must add up to 180. <7 + 36 = 180 so <7=144 so <1 is also 144.

7 0

2 years ago