Ask question

Ask question

All categories

3 years ago

5

Barry needs to find the area of a rectangular room with a length that is 2 feet longer than the width. Write an expression for t

he area of the rectangle in terms of the width.

1 answer:

babunello [35]3 years ago

5 0

Answer:

try asking a master, ok?

Step-by-step explanation:

You might be interested in

Would love help with this

Mars2501 [29]

It is always a rational number

-Hope This Helps-

6 0

3 years ago

The product of a number and 50

Daniel [21]

"Product" means we are looking at a multiplication problem.

Can be written as:

A number (n) times 50

n x 50

The simplest (and most common way) to write it:

50n

4 0

3 years ago

Read 2 more answers

I need some help with this.

zepelin [54]

The domain is the left side of the table, which is (x)

Range is the right side which is (y)

Your answer is B

4 0

3 years ago

please solve this and show your work, no links/ ip grabbers PLEASE i’ve been asking this forever i just need the answer

Lady bird [3.3K]

19x-15 = 15x -7
4x = 8
x =2
19(2) -15 = 23
15(2) -7 = 23

6 0

3 years ago

In the diagram below, assume that all points are given in rectangular coordinates. Determine the polar coordinates for each poin

Korolek [52]

Step-by-step explanation:

We have cartisean points. We are trying to find polar points.

We can find r by applying the pythagorean theorem to the x value and y values.

$r {}^{2} = {x}^{2} + {y}^{2}$

And to find theta, notice how a right triangle is created if we draw the base(the x value) and the height(y value). We also just found our r( hypotenuse) so ignore that. We know the opposite side and the adjacent side originally. so we can use the tangent function.

$\tan(x) = \frac{y}{x}$

Remeber since we are trying to find the angle measure, use inverse tan function

$\tan {}^{ - 1} ( \frac{y}{x} ) =$

Answers For 2,5

${2}^{2} + {5}^{2} = \sqrt{29} = 5.4$

So r=sqr root of 29

$\tan {}^{ - 1} ( \frac{5}{2} ) = 68$

So the answer is (sqr root of 29,68).

For -3,3

${ -3 }^{2} + {3}^{2} = \sqrt{18} = 3 \sqrt{2}$

$\tan {}^{ - 1} ( \frac{3}{ - 3} ) = - 45$

Use the identity

$\tan(x) = \tan(x + \pi)$

So that means

$\tan(x) = 135$

So our points are

(3 times sqr root of 2, 135)

For 5,-3.5

${5}^{2} + {3.5}^{2} = \sqrt{37.25}$

$\tan {}^{ - 1} ( \frac{ - 3.5}{ - 5} ) = 35$

So our points are (sqr root of 37.25, 35)

For (0,-5.4)

${0}^{2} + { - 5.4}^{2} = \sqrt{} 29.16 = 5.4$

So r=5.4

$\tan {}^{ - 1} (0) = undefined$

So our points are (5.4, undefined)

4 0

3 years ago