I NEED THIS PLEASE EXPERS.

Mathematics

2 answers:

kipiarov [429]2 years ago

7 0

Answer: The answer is 24 :)

Give brainlest if I’m right

melamori03 [73]2 years ago

6 0

Answer:

The answer I think it will be 24 but not sure.

Step-by-step explanation:

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Triangle ABC was dilated using the rule Dy,5/4

DIA [1.3K]

Answer:

10 units

Step-by-step explanation:

Just trust me

5 0

3 years ago

If f(x)=-3x-1, find f(-5) a)16 b)15 c)-16 d)14

Anuta_ua [19.1K]

Answer:

D) 14

Step-by-step explanation:

f(x) = -3x - 1

f(-5) = -3(-5) - 1

f(-5) = 15 - 1

f(-5) = 14

D) 14

4 0

3 years ago

Evaluate : limx→ tan2-sin2x x3

GrogVix [38]

Given:

$\lim _{x\to0}\frac{\tan 2x-\sin 2x}{x^3}$

Solve:

$\lim _{x\to0}\frac{\tan 2x-\sin 2x}{x^3}$

Use l'hopital's rule:

$\begin{gathered} =\lim _{x\to0}\frac{\frac{d}{dx}(-\sin 2x+\tan 2x)}{\frac{d}{dx}(x^3)} \\ =\lim _{x\to0}\frac{-2\cos (2x)+2\tan ^2(2x)+2}{3x^2} \end{gathered}$

Simplify:

$\begin{gathered} =\lim _{x\to0}\frac{-2\cos (2x)+2\tan ^2(2x)+2}{3x^2} \\ =\lim _{x\to0}\frac{2(-\cos (2x)+\tan ^2(2x)+1)}{3x^2} \end{gathered}$

Apply the constant multiple rule:

$\begin{gathered} \lim _{x\to0}cf(x)=c\lim _{x\to0}f(x) \\ \text{With c=}\frac{2}{3} \\ f(x)=\frac{-\cos (2x)+\tan ^2(2x)+1}{x^2} \end{gathered}$ $\begin{gathered} =\frac{2\lim _{x\to0}\frac{-\cos (2x)+\tan ^2(2x)+1}{x^2}}{3} \\ =\frac{2\lim _{x\rightarrow0}\frac{(4\tan ^2(2x)+4)\tan (2x)+2\sin (2x)}{2x}}{3} \end{gathered}$

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