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irga5000 [103]
4 years ago
6

($1.75+2times $0.25 +5times $0.05) times 24

Mathematics
2 answers:
Anastaziya [24]4 years ago
8 0
(1.75 + 2 x 0.25 + 5 x 0.05) x 24= 17
KengaRu [80]4 years ago
4 0

Answer:

60

Step-by-step explanation:1.75+(2x0.25=.50)=2.25+(5x0.05=0.25)=2.50x24=60


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If the work required to stretch a spring 2 ft beyond its natural length is 6 ft-lb, how much work is needed to stretch it 6 in.
mel-nik [20]

Answer:

0.375 feet-lb

Step-by-step explanation:

We have been given that the work required to stretch a spring 2 ft beyond its natural length is 6 ft-lb. We are asked to find the work needed to stretch the spring 6 in. beyond its natural length.

We can represent our given information as:

6=\int\limits^2_0 {F(x)} \, dx

We will use Hooke's Law to solve our given problem.

F(x)=kx

Substituting this value in our integral, we will get:

6=\int\limits^2_0 {kx} \, dx

Using power rule, we will get:

6=\left[ \frac{kx^2}{2} \right ]^2_0

6=\frac{k(2)^2}{2}-\frac{k(0)^2}{2}

6=\frac{4k}{2}-0\\\\k=3

We know that 6 inches is equal to 0.5 feet.

Work needed to stretch it beyond 6 inches beyond its natural length would be \int\limits^{0.5}_0 {kx} \, dx =\int\limits^{0.5}_0 {3x} \, dx

Using power rule, we will get:

\int\limits^{0.5}_0 {3x} \, dx = \left [\frac{3x^2}{2}\right]^{0.5}_0

\frac{3(0.5)^2}{2}-\frac{3(0)^2}{2}\Rightarrow \frac{3(0.25)}{2}-0=\frac{0.75}{2}=0.375

Therefore, 0.375 feet-lb work is needed to stretch it 6 in. beyond its natural length.

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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

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