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3 years ago

6

What is a equation????

2 answers:

8_murik_8 [283]3 years ago

5 0

answer: an equation is a set of numbers put together within a problem that you are meant to solve in order to receive the final answer.

Lelu [443]3 years ago

4 0

A statement that the values of two mathematical expressions are equal (indicated by the sign =).

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30men can finish 1/7of a work in 2days.18men will do othe remaining work in ?

Dovator [93]

7/7-1/7=6/7 is the rest of work
1/7 work takes 2days
6*1/7=6/7 work takes 6*2days=12days
30 men takes 12 days to finish the rest of work (6/7)
18 men takes 18*12/30= 7.2days

7 0

3 years ago

I really need help on this please and thank you in advance

sasho [114]

Answer:

Step-by-step explanation:

first, your answer will be negative,

x^12*(2y)^3=x^12(8y^3)

denominator= -x^2*y^12

after we divide them=- 8 x^10/y^9

choice E

7 0

3 years ago

the function of y=log(x) is translated 1 unit right and 2 units down. which is the graph of the translated function

faltersainse [42]

The answer is B. because its B.

4 0

3 years ago

Read 2 more answers

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

What percent is represented by the shaded area?<br>T

Fudgin [204]

There ain’t even T on the graph show the whole graph and maybe we can help you answer the question Karen

6 0

3 years ago