3 years ago

Need help with this one thanks in advance

Mathematics

2 answers:

Irina18 [472]3 years ago

8 0

D would most likely be the answer.

evablogger [386]3 years ago

7 0

The answer is 1 1/2 I simplified it

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DIA [1.3K]

Question 4. Just replace x with 2a.
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= 20a^2 + 8a

Question 5. Since x = 3, which is more than 1, use the second equation.
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3 years ago

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2/6 I hope this helps!

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3 years ago

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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:

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

Similary :

$\begin{gathered} =\frac{2\lim _{x\to0}(2\cos (2x)+12\tan ^4(2x)+16\tan ^2(2x)+4)}{3} \\ =\frac{2(6)}{3} \\ =4 \end{gathered}$

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1 year ago

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Lemur [1.5K]

Answer:

Where is the figure or picture

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3 years ago

78 is 15% of what number? Solve using an equation. Show your work.

Phoenix [80]

Answer:

520

Step-by-step explanation:

78 = 15% × (what number)

78/0.15 = (what number) = 520

78 is 15% of 520.

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3 years ago