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3 years ago

8

Find the midpoint of the segment with the given endpoints. (710) and (-7-3)

1 answer:

Annette [7]3 years ago

8 0

Answer:

717/2. 3/2

Step-by-step explanation:

(710+7)/2. (0--3)/2

717/2. 3/2

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Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

3 years ago

On Monday, 279 students went on a trip

belka [17]

45 students in each bus

7 0

3 years ago

A teenager who is 5 feet tall throws an object into the air. The quadratic function LaTeX: f\left(x\right)=-16x^2+64x+5f ( x ) =

tia_tia [17]

Answer:

At approximately x = 0.08 and x = 3.92.

Step-by-step explanation:

The height of the ball is modeled by the function:

$f(x)=-16x^2+64x+5$

Where f(x) is the height after x seconds.

We want to determine the time(s) when the ball is 10 feet in the air.

Therefore, we will set the function equal to 10 and solve for x:

$10=-16x^2+64x+5$

Subtracting 10 from both sides:

$-16x^2+64x-5=0$

For simplicity, divide both sides by -1:

$16x^2-64x+5=0$

We will use the quadratic formula. In this case a = 16, b = -64, and c = 5. Therefore:

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substitute:

$\displaystyle x=\frac{-(-64)\pm\sqrt{(-64)^2-4(16)(5)}}{2(16)}$

Evaluate:

$\displaystyle x=\frac{64\pm\sqrt{3776}}{32}$

Simplify the square root:

$\sqrt{3776}=\sqrt{64\cdot 59}=8\sqrt{59}$

Therefore:

$\displaystyle x=\frac{64\pm8\sqrt{59}}{32}$

Simplify:

$\displaystyle x=\frac{8\pm\sqrt{59}}{4}$

Approximate:

$\displaystyle x=\frac{8+\sqrt{59}}{4}\approx 3.92\text{ and } x=\frac{8-\sqrt{59}}{4}\approx0.08$

Therefore, the ball will reach a height of 10 feet at approximately x = 0.08 and x = 3.92.

7 0

2 years ago

Solve the equation -2(m-30)=-6m

Sergio039 [100]

Answer:

m=-15

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable. m=-15

7 0

3 years ago

Mobile users in India have gone up by 20% percent in a year. There are 540 million mobile users today.

DanielleElmas [232]

432 million, 540 times .20 equals 108, subtract 108 from 540 and you get your answer

7 0

3 years ago