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xeze [42]
4 years ago
6

Use the triangle below and the given trigonometric ratio to determine which angle in the triangle below is

Mathematics
1 answer:
4vir4ik [10]4 years ago
7 0
Cosine is the ratio of the adjacent side to the hypotenuse.

In short, we have this equation
cos(angle) = adjacent/hypotenuse

In the case of cos(A) = 18/19.5 we can see that
adjacent = 18
hypotenuse = 19.5

The side of 18 cm is next to angle 2, so the 18 cm side is adjacent to angle 2 (in contrast, the 7.5 cm side is the opposite leg in relation to angle 2)

So that's why the answer is angle 2
In other words, replace the '2' angle marker with 'A' to be able to write cos(A) = 18/19.5
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What is the value of the expression 6z2+2z+10 when z=3?
grigory [225]

Hi there! Hope this helps, and if it does, please mark brainliest!

Answer:

346

Step-by-step explanation:

6z2+2z+10

(6 x 3)^2 + 2 x 3 + 10

18^2 + 12 + 10

324 + 22

346

5 0
4 years ago
Solve each equation.<br> a. 6x + 4 = 34
Natasha_Volkova [10]
6x=30
x=5
the answer is 5
7 0
3 years ago
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What is the slope of the line y<br> 5<br> X-5?
nata0808 [166]

Answer:

the slope is 5

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8 0
3 years ago
How can you check if a point is an element of the function?
choli [55]

Answer:

Graph or make a table of the function.

Step-by-step explanation:

When you graph the function just look on the graph to see that point is part of your function. Same would go for the table. Just look and see if your numbers match.

6 0
3 years ago
First derivative of <br>√{cosec2x).show with full step.​
Mice21 [21]

Answer:

- \sf \displaystyle \:   \frac{ \cos(2x) }{ \sin ^{2} (2x)\sqrt{ \csc(2x) } }

Step-by-step explanation:

we are given a derivative

\displaystyle \:  \frac{d}{dx} ( \sqrt{  \csc(2x) } )

and said to figure out the first derivative

to do so

recall chain rule:

\sf\displaystyle \:  \frac{d}{dx} (f(g(x)) =  \frac{d}{dg} (f(g(x)) \times  \frac{d}{dx} (g)

so we get

\displaystyle \: g(x) =  \csc(2x)

rewrite the derivative using the chain rule:

\displaystyle \:  \frac{d}{dg} ( \sqrt{  g } )  \times  \frac{d}{dx} ( \csc(2x) )

use square root derivative rule to simplify:

\displaystyle \:   \frac{1}{ 2\sqrt{g} }  \times  \frac{d}{dx} ( \csc(2x) )

now we need to again use chain rule composite function derivative to simplify

where we'll take a new function n so we won't mess up two g's and we'll take 2x as n

use composite function derivative to simplify:

\sf \displaystyle \:   \frac{1}{ 2\sqrt{g} }  \times  \frac{d}{dn}( \csc(n) ) \times  \frac{d}{dx} (2x)

use derivative formula to simplify derivatives:

\sf \displaystyle \:   \frac{1}{ 2\sqrt{g} }  \times   - \cot(n)   \csc(n)  \times  2

substitute the value of n:

\sf \displaystyle \:   \frac{1}{ 2\sqrt{g} }  \times   - 2\cot(2x)   \csc(2x)

substitute the value of g:

\sf \displaystyle \:   \frac{1}{ 2\sqrt{ \csc(2x) } }  \times   - 2\cot(2x)   \csc(2x)

now we need our trigonometric skills to simplify

rewrite cot and csc:

\sf \displaystyle \:   \frac{1}{ 2\sqrt{ \csc(2x) } }  \times   - 2 \dfrac{ \cos(2x) }{ \sin(2x) }   \dfrac{1}{ \sin(2x) }

simplify multiplication:

\sf \displaystyle \:   \frac{1}{ \cancel{ \:  2}\sqrt{ \csc(2x) } }  \times    \cancel{- 2} \dfrac{ \cos(2x) }{ \sin ^{2} (2x) }

simplify multiplication:

- \sf \displaystyle \:   \frac{ \cos(2x) }{ \sin ^{2} (2x)\sqrt{ \csc(2x) } }

4 0
3 years ago
Read 2 more answers
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