Ask question

Ask question

All categories

Dmitry_Shevchenko [17]

2 years ago

15

I don’t understand what i can do

1 answer:

alexandr402 [8]2 years ago

4 0

Answer:

14.037

Step-by-step explanation:

$3^{-3} +5+9\\$

0.037+5+9

14.037

You might be interested in

Pls help with these questions, will give BRAINLIEST (if you know the answer)

Anika [276]

I don't know but these can be really difficult

good luck

8 0

2 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

2 years ago

What is the simplest form of 4 square root 81x^8y^5?

olasank [31]

Firstly,
thanks for posting this.

81 = 9x9 or 9 squared so at this stage I'm guessing the last two.
x^ 8 = (x^ 4) x (x^ 2) (its called the multiplication rule of indices).
Okay I have approx 14 mins to answer.
No panic
√ 81.x^8 x <span>√ y^5
</span>√ (9x^2 x 9x^6) x <span>√ y^5
</span>= 9x x <span>√ 9x^6 x </span><span>√ y^5
</span>did I just gain time?
√ 9x^6 = <span>√ </span>3x ^3 x <span>√ 3x ^ 3
</span>or 2 . √ 3x^3. <span>√ y^5
</span>more time
okay, so that wasnt successful.
√ 9x^6 =<span>√ </span> (3x^2) x <span>√ </span>(3x^4)

5 0

3 years ago

What is the value of 3 in 384?<br> Please explain?!

seropon [69]

Answer:uhhhhh 300

Step-by-step explanation:

I don’t really need to explain bc all u need to do is take the 80 and 4 away and u have only 300

3 0

2 years ago

How do I solve the linear equation of 27=6(x)+4(y) over 20=2(x)+5(y)

larisa86 [58]

<u>27 = 6x + 4y </u>= 1 7/20 = 3 4/5<u>
</u>20 = 2x + 5y<u>
</u>

8 0

3 years ago