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3 years ago

Help will mark you Brainliest

Mathematics

1 answer:

Ymorist [56]3 years ago

8 0

Answer:

6.6 inches

Step-by-step explanation:

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Solve for x. help please lol

Readme [11.4K]

Maybe you should try 18 it might work because it will all be equal

7 0

3 years ago

Read 2 more answers

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.

3 0

3 years ago

Please help ASAP ........ Find the area of the region bounded by the curves

guajiro [1.7K]

Answer:

(x₁ , y₁) = (0,0)

(x₂ , y₂) = $\frac{\sqrt[16]{3} }{19}$ , $\frac{12}{19}$

6 0

4 years ago

15 × 3 3/5 = _ +_= plz answer quickly please

Oksana_A [137]

Answer:

Step-by-step explanation:

15/1 × 3 3/5 = 54

~~~~~~~~~~

3 0

4 years ago

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia and older subjects before treatment 14

RideAnS [48]

Answer:

99% confidence interval estimate of the mean weight time for a population with drug treatments

(79.11 , 112.69)

Step-by-step explanation:

Step(i):-

Given data mean of the Population = 104 minutes

Given sample size 'n' =14

Mean of the sample(x⁻) = 95.9 minutes

Given Standard deviation = 24.4 minutes

Step(ii):-

99% confidence interval estimate of the mean weight time for a population is determined by

$(x^{-} - Z_{0.01} \frac{S.D}{\sqrt{n} } , x^{-} +Z_{0.01} \frac{S.D}{\sqrt{n} })$

$(95.9 - 2.576\frac{24.4}{\sqrt{14} } , 95.9+2.576\frac{24.4}{\sqrt{14} })$

On calculation , we get

(95.9 -16.79 , 95.9 +16.79)

(79.11 , 112.69)

Conclusion:-

99% confidence interval estimate of the mean weight time for a population with drug treatments

(79.11 , 112.69)

7 0

3 years ago

Help will mark you Brainliest​

Help will mark you Brainliest