Ask question

Ask question

All categories

4 years ago

11

Please help with the chart?!?!!?!!?!!?!!?!??!!?!

1 answer:

Elanso [62]4 years ago

6 0

Answer:

I'm sorry, I can't really see most of the chart in order to help you. Could you put a zoomed out one instead?

Step-by-step explanation:

The point is to find the average difference or generally "how far each are"

You might be interested in

Friends, i need help with this question.

Ratling [72]

Answer:

Step-by-step explanation:

The answer is 4.

Once you add 4, you get:

x^2 -2x + 4 = 7.

The left side is factorable:

(x-2)^2 = 7.

There is your perfect square.

6 0

3 years ago

I need help and thanks

EleoNora [17]

S+c=1.59 You need to add the savings and the sale price together.

8 0

3 years ago

What is the midpoint of (-1,5) (5,5)

mafiozo [28]

(-1,5)(5,5)
midpoint formula : (x1 + x2)/2, (y1+ y2)/2

(-1 + 5)/2 , (5 + 5)/2) =
(4/2 , 10/2) =
(2,5) <===

5 0

3 years ago

1) Lisa took a trip to Kuwait. upon leaving she decided to convert all of her dinars back into dollars . how many dollars did sh

Yuliya22 [10]

Lisa took a trip to Kuwait upon leaving she decided to convert all of her dinars back into dollars how many

3 0

3 years ago

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