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3 years ago

9

What is the missing reason in Step 8?

1 answer:

Lynna [10]3 years ago

4 0

The missing reason is you didn't put the actual question on here.

please show the full question.

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Help me Please!! <br><br> Choose All Answers that describe the polygon ABCD if…

Firlakuza [10]

Answer

A, B, C, D, E, F.

Step-by-step explanation:

In all of these shapes, the description applies to all except the trapezoid.

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3 years ago

What is the explicit formula for the sequence? 6, 5, 4, 3, 2, . . .

GuDViN [60]

I think that the right answer for this question is 6

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3 years ago

Multiplying fraction what is 5/8x6/15

BigorU [14]

$\text {Hello! Let's Solve this Problem!}$

$\text {When multiplying fractions you multiply numerator by numerator and }$ $\text {demoninator by denominator. Simplifying is required when the answer comes}$ $\text {out as fraction that can be simplified. Simplify until it's reduced to the lowest}$ $\text {fraction it ca be.}$

$\text {\underline{Multiply}}$

$\text {5*6=30}$

$\text {8*15=120}$

$\text {Your Answer Would Be:}$

$\Huge\boxed {30/120}$

$\text {Since this is a fraction we can simplify we simplify it by 30}$

$\text {Q: What does simplify mean?}$

$\text {A: It means to divide. So if someone says simplify 2/4 that means divide.}$

$\text {\underline {Divide}}$

$\text {30/120 divided by 30 =}$

$\text {Your Simplified Fraction Answer Would Be:}$

$\Huge\boxed {1/4}$

$\text {Best of Luck!}$

3 0

3 years ago

Read 2 more answers

(3 x 7) + (3 x 5) is the same as

Lena [83]

678.987Answer:

Step-by-step explanation:

5 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