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8_murik_8 [283]
3 years ago
7

Is x greater than, less than, or equal to 110°?

Mathematics
2 answers:
artcher [175]3 years ago
8 0

Answer:

x is equal to 110°. ( being vertically opposite angles)

Svet_ta [14]3 years ago
7 0
Hey babes!

I’m certainly positive that x is equal to 110 degrees.

I hope this helps!
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In which of the expressions below will x = 12.5? Select all
erma4kov [3.2K]

The answer is D

2x / 5 = 5

5 x 5 = 25

2x = 25

25/2

x= 12.5

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3 years ago
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The tallest water slide is 49 meters tall, Phineas
Oliga [24]

Answer:

n=9

Step-by-step explanation:

when you add 49 and 5 you get 54 then you need to divide it by 6

6 0
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Use the properties of logarithms to expand the following expression as much as possible. Simplify any numerical expressions that
sweet [91]

Answer:

f(x,y) = \log_{4} (x-5-\sqrt{25-6\cdot y})+\log_{4} (x-5+\sqrt{25-6\cdot y})

Step-by-step explanation:

Let be f(x,y) = \log_{4}(2\cdot x^{2}-20\cdot x +12\cdot y), this expression is simplified by algebraic and trascendental means. As first step, the second order polynomial is simplified. Its roots are determined by the Quadratic Formula, that is to say:

r_{1,2} = \frac{20\pm \sqrt{(-20)^{2}-4\cdot (2)\cdot (12\cdot y)}}{2\cdot (2)}

r_{1,2} = 5\pm \sqrt{25-6\cdot y}

The polynomial in factorized form is:

(x-5-\sqrt{25-6\cdot y})\cdot (x-5+\sqrt{25-6\cdot y})

The function can be rewritten and simplified as follows:

f(x,y) = \log_{4} [(x-5-\sqrt{25-6\cdot y})\cdot (x-5+\sqrt{25-6\cdot y})]

f(x,y) = \log_{4} (x-5-\sqrt{25-6\cdot y})+\log_{4} (x-5+\sqrt{25-6\cdot y})

3 0
3 years ago
a drone costs 300 plus 25 for tax for each set of extra propellers. what is the cost of a drone and 4 extra set of propellers​
Kobotan [32]
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3 0
3 years ago
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Flagpole to the ground using a piece of rope that is 10 feet long. He is going to tie the rope to the top of the flagpole and th
Flauer [41]

Answer:

<h3>6 feet</h3>

Step-by-step explanation

Using the pythagoras theorem;

Given

The length of the flag pole = 8 feet = opposite side

Length of the rope = 10 feet = hypotenuse

To determine how far out on the ground he need to secure the rope from the flagpole so that the rope is tight, we need to look for the adjacent. Using the equation

hyp² = opp² + adj²

10² = 8² + adj²

adj² = 100-64

adj² = 36

adj = √36

adj = 6feet

Hence the rope should be placed 6feet out of the ground

4 0
3 years ago
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