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Igoryamba
2 years ago
8

In the figure below, find the value of cose:

Mathematics
1 answer:
MariettaO [177]2 years ago
4 0

Answer:

D

Step-by-step explanation:

cosΘ = \frac{adjacent}{hypotenuse} = \frac{x}{r}

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Use the above figure to answer the question. What's the angle between the tangents at T in the figure? A. 90° B. 75° C. 60° D. 3
morpeh [17]
A
as the horizontal line and.vertical line meet at 90°
6 0
3 years ago
When you divide a whole number by a decimal less than 1, the quotient is greater than the whole number. Why?
LuckyWell [14K]
1 can be interpreted as a "whole" thing, and smaller than 1 is a part of a thing.

imagine: if you have some cakes and you count the number of cake pieces, those will be more than the number of cakes: as each cake will have more than one piece! (otherwise it wouldn't be called a piece!)

so the result of a division by a fraction is bigger than the original number, there are more parts than wholes in the results, as each whole has multiple parts!

for example: divide 3 cakes into pieces which are 1/4th of a cake. how many such pieces would there be? 12! more than the 3 original cakes;

is this possible to understand? tell me if not!
8 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7Bx....%7D%20%7D%20%7D%20%7D%20%20%3D%20%
andrey2020 [161]

First observe that if a+b>0,

(a + b)^2 = a^2 + 2ab + b^2 \\\\ \implies a + b = \sqrt{a^2 + 2ab + b^2} = \sqrt{a^2 + ab + b(a + b)} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{\cdots}}}}

Let a=0 and b=x. It follows that

a+b = x = \sqrt{x \sqrt{x \sqrt{x \sqrt{\cdots}}}}

Now let b=1, so a^2+a=4x. Solving for a,

a^2 + a - 4x = 0 \implies a = \dfrac{-1 + \sqrt{1+16x}}2

which means

a+b = \dfrac{1 + \sqrt{1+16x}}2 = \sqrt{4x + \sqrt{4x + \sqrt{4x + \sqrt{\cdots}}}}

Now solve for x.

x = \dfrac{1 + \sqrt{1 + 16x}}2 \\\\ 2x = 1 + \sqrt{1 + 16x} \\\\ 2x - 1 = \sqrt{1 + 16x} \\\\ (2x-1)^2 = \left(\sqrt{1 + 16x}\right)^2

(note that we assume 2x-1\ge0)

4x^2 - 4x + 1 = 1 + 16x \\\\ 4x^2 - 20x = 0 \\\\ 4x (x - 5) = 0 \\\\ 4x = 0 \text{ or } x - 5 = 0 \\\\ \implies x = 0 \text{ or } \boxed{x = 5}

(we omit x=0 since 2\cdot0-1=-1\ge0 is not true)

3 0
1 year ago
Write a function in any form that would match the graph shown below.
disa [49]

Answer:

0, 5

Step-by-step explanation:

I am not shure that is right but I think an ordered pair is form of a funtion.

6 0
3 years ago
Triangle xyz is similar to triangle xvw. write the congruence statements that must be true.
Lunna [17]

Answer:

∠ Y = ∠ V

∠ Z = ∠ W

\frac{XY}{XV} =\frac{YZ}{VW} =\frac{XZ}{XW}

Step-by-step explanation:

It is given that Δ XYZ is similar to Δ XVW.

Since corresponding angles of similar triangles are equal,

∠ X = ∠ X

∠ Y = ∠ V

∠ Z = ∠ W

Also, since corresponding sides of similar triangles are in proportionate,

\frac{XY}{XV} =\frac{YZ}{VW} =\frac{XZ}{XW}

7 0
3 years ago
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