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alex41 [277]
3 years ago
7

A certain television is advertised as a 50-inch TV (the diagonal length). If the width of

Mathematics
1 answer:
Anvisha [2.4K]3 years ago
3 0

Hello there i hope you are having a great day :) Your question: A certain television is advertised as a 50-inch TV (the diagonal length). If the width of  the TV is 14 inches, how many inches tall is the TV?

So you would need to used the Pythagorean theorem that would be

c^2 = a^2 + b^2

So this would equal letter C as The hypotenuse of the right angle and also letter A and B are the stands on the triangle so C would equal 50 inches and A and B would equal 14 inches.

Then the Hard bit so you would be,

1) a^2 = c^2 - b^2

2) a = √c^2 - b^2

3) a =  50^2 - 14^2

4) a = √ 2,500 - 196

5) a = √2,304

6) A = 26

So the answer would be 26 Hopefully ❤

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lukranit [14]

Question 21

Let's complete the square

y = 3x^2 + 6x + 5

y-5 = 3x^2 + 6x

y - 5 = 3(x^2 + 2x)

y - 5 = 3(x^2 + 2x + 1 - 1)

y - 5 = 3(x^2+2x+1) - 3

y - 5 = 3(x+1)^2 - 3

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Answer: Choice D

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

Question 22

Through trial and error you should find that choice D is the answer

Basically you plug in each of the given answer choices and see which results in a true statement.

For instance, with choice A we have

y < -4(x+1)^2 - 3

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Choice D is the answer because

y < -4(x+1)^2 - 3

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which is true since -9 is to the left of -7 on the number line.

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

Question 25

Answer: Choice B

Explanation:

The quantity (x-4)^2 is always positive regardless of what you pick for x. This is because we are squaring the (x-4). Squaring a negative leads to a positive. Eg: (-4)^2 = 16

Adding on a positive to (x-4)^2 makes the result even more positive. Therefore (x-4)^2 + 1 > 0 is true for any real number x.

Visually this means all solutions of y > (x-4)^2 + 1 reside in quadrants 1 and 2, which are above the x axis.

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