All categories

3 years ago

If a= .75 and b= .50, what is the perimeter of the rectangle?

Mathematics

1 answer:

Lerok [7]3 years ago

3 0

Answer:

C. 14 inches

Step-by-step explanation:

The perimeter is double the sum of length and width:

P = 2((4a +3) +(6b -2) = 8a +12b +2

For the given values, this is ...

P = 8(0.75) +12(0.50) +2 = 6 + 6 + 2 = 14

The perimeter is 14 inches.

You might be interested in

What is 21 inches to 9 feet in simplest form

Olenka [21]

2.41667, you can also find a converter online.

7 0

3 years ago

Is 9/3 a whole number

Liono4ka [1.6K]

Yes! 3x3= 9. Therefore, 3 goes into nine 3 times. So the whole number would be 3.

6 0

3 years ago

Read 2 more answers

Three friends went apple picking and then counted the apples they picked.Who had the largest red to green apple ratio? Use compl

antoniya [11.8K]

Answer:

Carmen had the largest red to green apple ratio of 5 : 3

Step-by-step explanation:

Red apples : Green apples

Carmen = 20 red apples : 12 green apples

= 20 / 12

= 5 / 3 = 1.67

= 5 : 3

Dylan = 12 red apples : 10 green apples

= 12 / 10

= 6 / 5 = 1.2

= 6 : 5

Eli = 16 red apples : 20 green apples

= 16 / 20

= 4 / 5 = 0.8

= 4 : 5

Carmen had the largest red to green apple ratio of 5 : 3

7 0

2 years ago

The average age of 6 men is 35 years and

PSYCHO15rus [73]

Answer:

39.50

42.50

Step-by-step explanation:

Age of the 2 men = sum of the ages of the 6 men - sum of the ages of the 4 men

Average = sum of ages / total number of men

35 = sum of ages / 6

sum of ages = 35 x6 = 210

32 = sum of ages / 4

32 x 4 = sum of ages

128

210 - 128 = 82

age of the younger man = x

age of the older man = 3 + x

x + 3 + x = 82

2x + 3 = 82

2x = 82 - 3

2x = 79

x = 79/2 = 39.5

older man = 39.5 + 3 = 42.50

7 0

2 years ago

Determine whether the improper integral converges or diverges, and find the value of each that converges.

Ksju [112]

Answer:converge at $I=\frac{1}{3}$

Step-by-step explanation:

Given

Improper Integral I is given as

$I=\int^{\infty}_{3}\frac{1}{x^2}dx$

integration of $\frac{1}{x^2}$ is - $\frac{1}{x}$

$I=\left [ -\frac{1}{x}\right ]^{\infty}_3$

substituting value

$I=-\left [ \frac{1}{\infty }-\frac{1}{3}\right ]$

$I=-\left [ 0-\frac{1}{3}\right ]$

$I=\frac{1}{3}$

so the value of integral converges at $\frac{1}{3}$

8 0

3 years ago