Mathematics: Marvin buys a hose that is 27 3/4 feet long he already owns a hose at home that is 2/3 the length of the new hose how many total yards of hose does marvin have now?

Marvin buys a hose that is 27 3/4 feet long he already

Ответ:

DanteMadrid33

06.01.2022

Mathematics

The total yards of hose that Marvin has now is $15\frac{5}{12}$ yards

Further Explanation

New hose (x) : $27\frac{3}{4} feet$

at home hose (y) = $\frac{2}{3} x$

How many total yards of hose does Marvin have now?

First, we are going to calculate the long of the hose that Marvin already has:

$\boxed {y = \frac{2}{3} x }\\\boxed { = \frac{2}{3} \times \frac{111}{4} = \frac{222}{12} = 18 \frac{1}{2} }$

Total long of hose:

= new hose + at home hose

= x + y

$\boxed {= 27\frac{3}{4} + 18\frac{1}{2} }$

$\boxed {= 27+18 (\frac{3}{4} + \frac{1}{2}) }\\\boxed {= 45 (\frac{3+2}{4}) }\\\boxed {= 45\frac{5}{4} }\\ \boxed {= 46\frac{1}{4} }$

Total of the long hose Marvin has is $46\frac{1}{4}$ feet. The question asking the length in yards, so we need to convert feet to yards:

$1 foot = \frac{1}{3} yard$

$\boxed {= 46\frac{1}{4} feet \times\frac{1}{3} yard } \\\boxed {= \frac{185}{4} \times\frac{1}{3} }\\\boxed {= \frac{185}{12} = 15\frac{5}{12} }$

So the total yards of hose that Marvin has now is $15\frac{5}{12}$ yards

Learn More

Fraction subtraction

Mixed fraction (part to whole relationship)

Mixed fraction

Keywords: mixed fraction, improper fraction, additional mixed fraction

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Ответ:

potatogirl9300

01.02.2022

Mathematics

Step-by-step explanation:

h(t) = (t + 3)² + 5

A) h(t) = (t + 3)² + 5 -2 ≤ t ≤ 0

average rate of change in h over the change in t is given as:

$\frac{h(0)-h(-2)}{0-(-2)}=\frac{(0+3)^2+5-[(-2+3)^2+5]}{0+2}=\frac{14-6}{2} =4$

It does not have a negative average rate of change

B) A) h(t) = (t + 3)² + 5 -4 ≤ t ≤ -3

average rate of change in h over the change in t is given as:

$\frac{h(-3)-h(-4)}{-3-(-4)}=\frac{(-3+3)^2+5-[(-4+3)^2+5]}{-3+4}=\frac{5-6}{1} =-1$

It have a negative average rate of change

c) A) h(t) = (t + 3)² + 5 -3 ≤ t ≤ 4

average rate of change in h over the change in t is given as:

$\frac{h(4)-h(-3)}{4-(-3)}=\frac{(4+3)^2+5-[(-3+3)^2+5]}{4+3}=\frac{54-5}{7} =7$

It does not have a negative average rate of change

D) h(t) = (t + 3)² + 5 1 ≤ t ≤ 4

average rate of change in h over the change in t is given as:

$\frac{h(4)-h(1)}{4-(1)}=\frac{(4+3)^2+5-[(1+3)^2+5]}{4-1}=\frac{54-21}{3} =11$

It does not have a negative average rate of change

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Marvin buys a hose that is 27 3/4 feet long he already owns a hose at home that is 2/3 the length of the new hose how many total yards of hose does marvin have now?

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