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kykrilka [37]
4 years ago
8

Which expression is equivalent to 8(a-6) a. w>36 b w<36 c.w>5 d.w<5

Mathematics
1 answer:
beks73 [17]4 years ago
3 0

Answer:

8a - 48

Step-by-step explanation:

An equivalent expression is an expression which is equal to 8(a-6). You can expand or simplify an expression to make an equivalent expression. Apply the distributive property to form a new equivalent expression by multiplying 8 into each term. 8(a-6) = 8*a - 8*6 = 8a - 48. The options listed below do not connect to this problem since they do not use the same variable as the expression.

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(Factorize): 10- 10-1-3​
Wittaler [7]

Answer:

=−4

Step-by-step explanation:

10−10−1−3

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2 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) } }

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Are the triangles similar? If so, explain why.<br> АД<br> 3462
Alchen [17]

<em>Greetings from Brasil...</em>

Just 2 equal angles are enough for the triangles to be similar .... The second triangle has the same angles as the first.

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pishuonlain [190]

Answer:

808 over 25 as a faraction

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