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3 years ago

The graph shows equivalent ratios. Frank plotted another point with a y-value of 18. What is the x-value of that point?

Mathematics

1 answer:

barxatty [35]3 years ago

7 0

Answer:

the answer is 6 purrr

Step-by-step explanation:

i took the assignment

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In ΔUVW, the measure of ∠W=90°, the measure of ∠V=71°, and VW = 55 feet. Find the length of UV to the nearest tenth of a foot.

Katarina [22]

Answer:

168.9 feet

Step-by-step explanation:

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3 years ago

(x+iy)^4 -(x+iy)^2 +16=0

viktelen [127]

Answer:

below

Step-by-step explanation:

that is the procedure above

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3 years ago

Find 55% of 164 I don't get it I need help can someone help me like right now please

IgorC [24]

$55\%\ of\ 164\ is...\\\\ \frac{55}{100} \times \frac{164}{1} =\\\\55*164=9020\\100*1=100\\\\ \frac{9020}{100}\approx90\\\\20\ is\ left\ over.\\\\90.20\\\\\\55\%\ of\ 164\ is\ 90.2$

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4 years ago

Look at the image. (calculus)

mash [69]

<h3>Answer: Choice H) 2</h3>

=============================================

Explanation:

Recall that the pythagorean trig identity is $\sin^2 x + \cos^2x = 1$

If we were to isolate sine, then,

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

We don't have to worry about the plus minus because sine is positive when 0 < x < pi/2.

Through similar calculations, $\cos x = \sqrt{1-\sin^2x}\\\\$

Cosine is also positive in this quadrant.

-------------

So,

$\frac{\sqrt{1-\cos^2x}}{\sin x}+\frac{\sqrt{1-\sin^2x}}{\cos x}\\\\\frac{\sin x}{\sin x}+\frac{\cos x}{\cos x}\\\\1+1\\\\2$

Therefore,

$\frac{\sqrt{1-\cos^2x}}{\sin x}+\frac{\sqrt{1-\sin^2x}}{\cos x}=2$

is an identity as long as 0 < x < pi/2

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2 years ago

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Consider the linear function that is represented by the equation y = negative 10 x + 6 and the linear function that is represent

Vilka [71]

Answer: The function that is represented by the equation y = negative 10 x + 6 has a steeper slope and a greater y-intercept.

Step-by-step explanation: I took the unit test on edge hoped this helped <3

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3 years ago

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