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umka2103 [35]
3 years ago
10

In a square, the sum of the lengths of three sides is 45 and in a rectangle the difference between a length and a width is 5 in.

if a width of the rectangle is the same as a side of the square, how long is a length of the rectangle?
Mathematics
1 answer:
stira [4]3 years ago
5 0
Square:
We know all sides of a squares are equal in length.
s= the length of one side of the square.
45= s+s+s
45= 3s
s=15 inches
Each side of the square is 15 inches.

Rectangle:
length(l)= ????
Width(w)= s = 15 inches

l -w= 5inches
l- (15 inches)= 5 inches
l = 20 inches


Answer:
A length of the rectangle is 20 inches long.
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5 0
3 years ago
Which methods correctly solve for the variable n in the equation 11−n=−20? Select all that apply. First, add n to both sides. Th
Zina [86]

Answer:

First, subtract 11 from both sides. Then, multiply both sides by −1.

First, add n to both sides. Then, add 20 to both sides.

Step-by-step explanation:

Given equation:

11 - n = -20

Solve for variable n

11 - n = -20

Subtract 11 from both sides

11 - n - 11 = -20 - 11

- n = - 31

Multiply both sides by -1

- n(-1) = - 31(-1)

n = 31

First, subtract 11 from both sides. Then, multiply both sides by −1.

11 - n = -20

11 -n +n = -20 + n

11 = -20 + n

11 + 20 = -20 + n +20

31 = n

First, add n to both sides. Then, add 20 to both sides.

First, subtract n from both sides. Then, multiply both sides by −20.

First, add n to both sides. Then, subtract 11 from both sides.

11 - n = -20

11 - n +n = -20 + n

11 = -20 + n

11 - 11 = -20 + n -11

Not possible

First, subtract 11 from both sides. Then, multiply both sides by −1.

First, add n to both sides. Then, subtract 20 from both sides.

11 - n = -20

11 - n + n = -20 + n

11 = -20 + n

11 - 20 = -20 + n - 20

9 = n - 40

Not possible

5 0
3 years ago
The region bounded by y=x^2+1, y=x, x=-1, x=2 with square cross sections perpendicular to the x-axis.
VLD [36.1K]

Answer:

The bounded area is 5 + 5/6 square units. (or 35/6 square units)

Step-by-step explanation:

Suppose we want to find the area bounded by two functions f(x) and g(x) in a given interval (x1, x2)

Such that f(x) > g(x) in the given interval.

This area then can be calculated as the integral between x1 and x2 for f(x) - g(x).

We want to find the area bounded by:

f(x) = y = x^2 + 1

g(x) = y = x

x = -1

x = 2

To find this area, we need to f(x) - g(x) between x = -1 and x = 2

This is:

\int\limits^2_{-1} {(f(x) - g(x))} \, dx

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx

We know that:

\int\limits^{}_{} {x} \, dx = \frac{x^2}{2}

\int\limits^{}_{} {1} \, dx = x

\int\limits^{}_{} {x^2} \, dx = \frac{x^3}{3}

Then our integral is:

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx = (\frac{2^3}{2}  + 2 - \frac{2^2}{2}) - (\frac{(-1)^3}{3}  + (-1) - \frac{(-1)^2}{2}  )

The right side is equal to:

(4 + 2 - 2) - ( -1/3 - 1 - 1/2) = 4 + 1/3 + 1 + 1/2 = 5 + 2/6 + 3/6 = 5 + 5/6

The bounded area is 5 + 5/6 square units.

3 0
2 years ago
Solve the system by substitution.<br>3x-6y=2<br>4x+3y=-1​
crimeas [40]

Answer:

y = -1/3

x = 0

Step-by-step explanation:

3x-6y = 2    then 3x = 2+6y,   x=(2+6y)/3

4x+3y = -1

substitute for x

4(2/3+2y)-6y = 2

8/3+8y-6y = 2

reduce

2y = 2-8/3

2y = -2/3

divide both sides by 2

y = -1/3

4x+3(-1/3) = -1

4x = 0

x = 0

8 0
3 years ago
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