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3 years ago

A trampoline has a jumping surface that is 15.7 ft. long and 12 ft. wide. What is the area of the jumping surface?

Mathematics

2 answers:

Arisa [49]3 years ago

5 0

Answer:

188.4

Step-by-step explanation:

because if you multiply 15.7 and 12 you get 188.4 and that is the area of the jumping space.

allochka39001 [22]3 years ago

4 0

Answer:

area = length×wide=15.7×12=188.4ft²

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Consider the function f(x)=-2/3x +5 what is f(-6) enter your answer in the box f(-6)=

GaryK [48]

Answer:

f(- 6) = 9

Step-by-step explanation:

to evaluate f(- 6), substitute x = - 6 into f(x)

f(- 6) = - $\frac{2}{3}$ × - 6 + 5

= - $\frac{2(-6)}{3}$ + 5 = 4 + 5 = 9

4 0

3 years ago

Lisa wants to frame her little brothers drawing as a gift to her mother.the rectangle drawing is 43.5 centimeters by 934 millime

konstantin123 [22]

Answer:

Lisa will need $273.8\ cm$ of wood framing

Step-by-step explanation:

we know that

The perimeter of rectangle is equal to

$P=2(b+h)$

In this problem we have

$b=43.5\ cm$

$h=934\ mm$

convert to cm

Remember that $1\ cm=10\ mm$

$h=934\ mm=934/10=93.4\ cm$

substitute the values and solve the perimeter

$P=2(43.5+93.4)=273.8\ cm$

7 0

3 years ago

Lim t^4 - 6 / 2t^2 - 3t + 7

Harman [31]

I think you meant to say

$\displaystyle \lim_{t\to2}\frac{t^4-6}{2t^2-3t+7}$

(as opposed to x approaching 2)

Since both the numerator and denominator are continuous at t = 2, the limit of the ratio is equal to a ratio of limits. In other words, the limit operator distributes over the quotient:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)}$

Because these expressions are continuous at t = 2, we can compute the limits by evaluating the limands directly at 2:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)} = \frac{2^4-6}{2\cdot2^2-3\cdot2+7} = \boxed{\frac{10}9}$