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2 years ago

The area of a rectangle is 52 in^2. If the length is 9 inches more than the width, find the

Mathematics

2 answers:

7nadin3 [17]2 years ago

6 0

Answer: 52 divided by 9 which equals to 5+7/9 inches

Step-by-step explanation:

Bingel [31]2 years ago

4 0

Answer:

width=4in.

length=13in

Step-by-step explanation:

let the width of the rectangle be x

then the length will be x+9

Area= length×width

52= (x+9)x

52= x²+9x

x²+9x-52=0

x²+13x-4x-52=0

x(x+13) - 4(x+13)= 0

(x-4)(x+13)=0

x=4 or -13

dimension cannot be negative

hence x=4in

the width is 4in. and the length is 4+9=13in.

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Simplify 6 - 3(4 - 3x)<br> -3x - 6<br> 9x + 18<br> 9x - 6

svet-max [94.6K]

Answer:

6 (19 x - 2)

Step-by-step explanation:

Simplify the following:

6 - 3 (4 - 3 x) - 3 x - 6×9 x + 18×9 x - 6

-6×9 = -54:

6 - 3 (4 - 3 x) - 3 x + -54 x + 18×9 x - 6

18×9 = 162:

6 - 3 (4 - 3 x) - 3 x - 54 x + 162 x - 6

Grouping like terms, 6 - 3 (4 - 3 x) - 3 x - 54 x + 162 x - 6 = (-3 x - 54 x + 162 x) - 3 (4 - 3 x) + (6 - 6):

(-3 x - 54 x + 162 x) - 3 (4 - 3 x) + (6 - 6)

-3 x - 54 x + 162 x = 105 x:

105 x - 3 (4 - 3 x) + (6 - 6)

6 - 6 = 0:

105 x - 3 (4 - 3 x)

-3 (4 - 3 x) = 9 x - 12:

9 x - 12 + 105 x

Grouping like terms, 105 x + 9 x - 12 = (105 x + 9 x) - 12:

(105 x + 9 x) - 12

105 x + 9 x = 114 x:

114 x - 12

Factor 6 out of 114 x - 12:

Answer: 6 (19 x - 2)

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3 years ago

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Nitella [24]

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3 years ago

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Paula traveled 65 miles per hour for 2.5 hours and then traveled 45 miles per hour for 3.5 hours what is the total distance in m

Naddik [55]

Answer:

320

Step-by-step explanation:

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3 years ago

A Mexican restaurant sells quesadillas in two sizes: a "large" 10 inch-round quesadilla and a "small" 5 inch-

elena-14-01-66 [18.8K]

Answer:

Half of the 10-inch quesadilla is greater than the entire 5-inch quesadilla

Step-by-step explanation:

we know that

The area of a circle is equal to

$A=\pi r^2$

where

r is the radius

step 1

Find the area of the 10 inch-round quesadilla

we have

$D=10\ in$

$r=10/2=5\ in$

substitute

$A=\pi (5)^2$

$A=25\pi\ in^2$

step 2

Find the area of the 5 inch-round quesadilla

we have

$D=5\ in$

$r=5/2=2.5\ in$

substitute

$A=\pi (2.5)^2$

$A=6.25\pi\ in^2$

step 3

Which is larger, half of the 10-inch quesadilla or the entire 5-inch quesadilla?

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the entire 5-inch quesadilla ----> $6.25\pi\ in^2$

therefore

Half of the 10-inch quesadilla is greater than the entire 5-inch quesadilla

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago