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Alexandra [31]
3 years ago
8

PLS HELP WILL MARK BRAINLIEST Represent the following algebraically

Mathematics
1 answer:
Vesna [10]3 years ago
7 0

Answer:

width = (21 - 2n) ÷ 2 or 21/2 - n

Step-by-step explanation:

To find Width

let's say width is x

so 2x + 2n = 21

take away 2n from both sides

then 2x = 21 - 2n

Divide both sides by 2 to get x on its own

x = 21 - 2n /2

therefore the width of the rectangle is 21 - 2n /2

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Select the correct answer.
svet-max [94.6K]

Answer:

Step-by-step explanation:

A parallelogram is a 4-sided polygon whose opposite sides are equal. By definition, opposite angles are also equal. A square, rectangle, or rhombus are considered as a parallelogram. Among the choices, the correct answer is D. "Opposite angles are congruent"

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4 years ago
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The intercepts of the equation x^2 + 4y^2 = 16 are ___<br> PLEASE HELP
Eva8 [605]
Yintercepts are where x=0
set to zero to solve

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divide 4
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3 0
4 years ago
Question<br> Translate and solve: Twelve less than x is equal to 51. Enter the equation
mafiozo [28]

Answer:

x = 63

Step-by-step explanation:

From the question, we can deduce the following points;

- 12 less than x which simply means 12 subtracted from x.

- the value of the above expression is equal to 51.

Translating the word problem into an algebraic expression, we have;

x - 12 = 51

Rearranging the equation (collecting like terms), we have;

x = 51 + 12

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6 0
3 years ago
The radius of a circle is decreasing at a rate of 6.56.56, point, 5 meters per minute. At a certain instant, the radius is 12121
Ilia_Sergeevich [38]

Answer:

The rate of change of the area of the circle is  approximately -490.09 \ m^2/min.

Step-by-step explanation:

Given:

\frac{dr}{dt} = -6.5 \ m/min

radius r =12 \ m

We need to find the rate of change of the area of the circle at that instant.

Solution:

Now we know that;

Area of the circle is given by π times square of the radius.

framing in equation form we get;

A= \pi r^2

to find the rate of change of the area of the circle at that instant we need to take the derivative on both side.

\frac{dA}{dt}=\frac{d(\pi r^2)}{dt}\\\\\frac{dA}{dt}= 2\pi r \frac{dr}{dt}

Substituting the given values we get;

\frac{dA}{dt}= 2\times \pi \times 12 \times -6.5\\\\\frac{dA}{dt}\approx -490.09 \ m^2/min

Hence The rate of change of the area of the circle is  approximately -490.09 \ m^2/min.

6 0
3 years ago
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