Write an equation that represents a graphed linear function

PLEASE HELPPPPPP The perimeter is 50 ft and the length is 15 ft. What's the width?

SVEN [57.7K]

Answer:

10

Step-by-step explanation:

I assume it's a rectangle, therefore:

P = 50

a = 15 ft

b = ?

P = 2a + 2b

50 = 2 * 15 + 2b

50 = 30 + 2b

2b = 50 - 30

2b = 20

b = 10

8 0

3 years ago

Can you draw trapezoid three right angles

Hunter-Best [27]

Its impossible to draw a trapezoid with just three right angles.

a trapezoid has 4 sides, which means all the angles inside the trapezoid must add up to 360 degrees.

if you have just 3 right angles (90x3), you already use up 270 degrees. Leaving you with just 90 degrees left, which is also a right angle. That means, there has to be four, if you have at least 3.

6 0

4 years ago

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}$}$

6 0

2 years ago

Select the correct answer. Which of the following is a solution to ? A. B. C. D. E.

Leni [432]

Answer:

the answer to your question is A

3 0

3 years ago

Read 2 more answers

5. If position of object x = 3 sinΘ – 7 cosΘ then motion of object is bounded between position.

lesya692 [45]

9514 1404 393

Answer:

±√58 ≈ ±7.616

Step-by-step explanation:

The linear combination of sine and cosine functions will have an amplitude that is the root of the sum of the squares of the individual amplitudes.

|x| = √(3² +7²) = √58

The motion is bounded between positions ±√58.

_____

Here's a way to get to the relation used above.

The sine of the sum of angles is given by ...

sin(θ+c) = sin(θ)cos(c) +cos(θ)sin(c)

If this is multiplied by some amplitude A, then we have ...

A·sin(θ+c) = A·sin(θ)cos(c) +A·cos(θ)sin(c)

Comparing this to the given expression, we find ...

A·cos(c) = 3 and A·sin(c) = -7

We know that sin²+cos² = 1, so the sum of the squares of these values is ...

(A·cos(c))² +(A·sin(c))² = A²(cos(c)² +sin(c)²) = A²(1) = A²

That is, A² = (3)² +(-7)² = 9+49 = 58. This tells us the position function can be written as ...

x = A·sin(θ +c) . . . . for some angle c

x = (√58)sin(θ +c)

This has the bounds ±√58.

3 0

3 years ago