Can someone pls help me?

Mathematics

1 answer:

Setler [38]2 years ago

4 0

I believe the answer is AD=7.0

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Need help will give brainliest

Komok [63]

Answer:

40.56

Step-by-step explanation:

2pir

8÷2=r=4

2pi4=25.13

25.13÷2=12.565

10+10+8+12.56=40.565

6 0

3 years ago

So we know when x = 0, y = 3. What does y equal in the equation y=x2 + 3 if x = 1?

const2013 [10]

Answer:This line contains no two such points. Hence the slope is not defined. A y intercept occurs when we make x equal to 0 . Since this is prohibited by the equation x=−3 , there is no y intercept

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Look at the picture

Sonbull [250]

$\large\displaystyle\text{$\begin{gathered}\sf 9|x-8| < 36 \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf Divide \ both \ sides \ by \ 9. \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \frac{9(|x-8|)}{9} < \frac{36}{9} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf |x-8| < 4 \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf Solve \ Absolute \ Value. \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf |x-8| < 4 \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf We \ know \ x-8 < 4 \ and \ x-8 > -4 \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf x-8 < 4 \ (Condition \ 1) \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf x-8+8 < 4+8 \ (Add \ 8 \ to \ both \ sides) \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf x < 12 \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf x-8 > -4 \ (Condition \ 2) \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf x-8+8 > -4+8 \ (Add \ 8 \ to \ both \ \ sides) \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf x > 4 \end{gathered}$}$

$\underline{\boldsymbol{\sf{Answer}}}$

$\boxed{\large\displaystyle\text{$\begin{gathered}\sf x < 12 \ and \ x > 4 \end{gathered}$} }$

$\large\displaystyle\text{$\begin{gathered}\sf Therefore,\bf{\underline{the \ correct \ option}} \ \end{gathered}$}\large\displaystyle\text{$\begin{gathered}\sf is \ \bf{\underline{"A"}}. \end{gathered}$}$