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3 years ago

12

The length of a rectangle is double its width. The perimeter of the rectangle is 36 feet. What is the area, in square

1 answer:

Inga [223]3 years ago

7 0

Answer:

72 square feet

Step-by-step explanation:

In the attached file

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Use the remainder theorem to determine the remainder when d4 + 2d2 + 5d − 10 is divided by d + 4.

MakcuM [25]

D have a nice day sir/maam

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3 years ago

Which fraction is equal to one third

KIM [24]

Answer:

2/6, 3/9, 4/12 ...

Step-by-step explanation:

We can find equivalent fractions to 1/3 by multiplying the numerator, or top number, and the denominator, or bottom number, by the same factor.

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3 years ago

[-x - 2y = 7<br> [3x + y = 6

ZanzabumX [31]

Answer:

$multipling \: eq \: by \: 3 \\ then \: we \: get \: - 3x - 6y = 21 \\ now \: add \: both \: eq \\ - 5y = 27 \\ y = - 5.4$

Step-by-step explanation:

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8 0

3 years ago

Plot the point whose polar coordinates are given. Then find two other pairs of polar coordinates of this point, one with r >

Ira Lisetskai [31]

Answer:

The other pairs are:

$(a)\ (2, \frac{5\pi}{6}) \to$ $(2, \frac{17\pi}{6})$ and $(-2, \frac{23\pi}{6})$

$(b)\ (1, -\frac{2\pi}{3}) \to$ $(1, \frac{4\pi}{3})$ and $(-1, \frac{7\pi}{3})$

$(c)\ (-1, \frac{5\pi}{4}) \to$ $(-1, \frac{3\pi}{4} )$ and $(1, \frac{7\pi}{4})$

See attachment for plots

Step-by-step explanation:

Given

$(a)\ (2, \frac{5\pi}{6})$

$(b)\ (1, -\frac{2\pi}{3})$

$(c)\ (-1, \frac{5\pi}{4})$

Solving (a): Plot a, b and c

See attachment for plots

Solving (b): Find other pairs for $r > 0$ and $r < 0$

The general rule is that:

The other points can be derived using

$(r, \theta) = (r, \theta + 2n\pi)$

and

$(r, \theta) = (-r, \theta + (2n + 1)\pi)$

Let $n =1$ ---- You can assume any value of n

So, we have:

$(r, \theta) = (r, \theta + 2n\pi)$

$(r, \theta) = (r, \theta + 2*1*\pi)$

$(r, \theta) = (r, \theta + 2\pi)$

$(r, \theta) = (-r, \theta + (2n + 1)\pi)$

$(r, \theta) = (-r, \theta + (2*1 + 1)\pi)$

$(r, \theta) = (-r, \theta + (2 + 1)\pi)$

$(r, \theta) = (-r, \theta + 3\pi)$

$(a)\ (2, \frac{5\pi}{6})$

$r = 2\ \ \ \ \theta = \frac{5\pi}{6}$

So, the pairs are:

$(r, \theta) = (r, \theta + 2\pi)$

$(2, \frac{5\pi}{6}) = (2, \frac{5\pi}{6} + 2\pi)$

Take LCM

$(2, \frac{5\pi}{6}) = (2, \frac{5\pi+12\pi}{6})$

$(2, \frac{5\pi}{6}) = (2, \frac{17\pi}{6})$

And

$(r, \theta) = (-r, \theta + 3\pi)$

$(2, \frac{5\pi}{6}) = (-2, \frac{5\pi}{6} + 3\pi)$

Take LCM

$(2, \frac{5\pi}{6}) = (-2, \frac{5\pi+18\pi}{6})$

$(2, \frac{5\pi}{6}) = (-2, \frac{23\pi}{6})$

The other pairs are:

$(2, \frac{17\pi}{6})$ and $(-2, \frac{23\pi}{6})$

$(b)\ (1, -\frac{2\pi}{3})$

$r = 1\ \ \ \theta = -\frac{2\pi}{3}$

So, the pairs are:

$(r, \theta) = (r, \theta + 2\pi)$

$(1, -\frac{2\pi}{3}) = (1, -\frac{2\pi}{3} + 2\pi)$

Take LCM

$(1, -\frac{2\pi}{3}) = (1, \frac{-2\pi+6\pi}{3})$

$(1, -\frac{2\pi}{3}) = (1, \frac{4\pi}{3})$

And

$(r, \theta) = (-r, \theta + 3\pi)$

$(1, -\frac{2\pi}{3}) = (-1, -\frac{2\pi}{3} + 3\pi)$

Take LCM

$(1, -\frac{2\pi}{3}) = (-1, \frac{-2\pi+9\pi}{3})$

$(1, -\frac{2\pi}{3}) = (-1, \frac{7\pi}{3})$

The other pairs are:

$(1, \frac{4\pi}{3})$ and $(-1, \frac{7\pi}{3})$

$(c)\ (-1, \frac{5\pi}{4})$

$r = -1 \ \ \ \ \theta = \frac{-5\pi}{4}$

So, the pairs are

$(r, \theta) = (r, \theta + 2\pi)$

$(-1, \frac{-5\pi}{4}) = (-1, \frac{-5\pi}{4} + 2\pi)$

Take LCM

$(-1, \frac{-5\pi}{4}) = (-1, \frac{-5\pi+8\pi}{4} )$

$(-1, \frac{-5\pi}{4}) = (-1, \frac{3\pi}{4} )$

And

$(r, \theta) = (-r, \theta + 3\pi)$

$(-1, \frac{-5\pi}{4}) = (-(-1), \frac{-5\pi}{4}+ 3\pi)$

Take LCM

$(-1, \frac{-5\pi}{4}) = (1, \frac{-5\pi+12\pi}{4})$

$(-1, \frac{-5\pi}{4}) = (1, \frac{7\pi}{4})$

So, the other pairs are:

$(-1, \frac{3\pi}{4} )$ and $(1, \frac{7\pi}{4})$

5 0

3 years ago

What is the volume of the right prism?

grigory [225]

V = 1/2B*H*L
V = 12*35*40/2
V= 16,800/2
V= 8,400 $in^{3}$

The volume of the right prism is B) 8,400 $in^{3}$

7 0

4 years ago

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