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3 years ago

What is 8.023 in expanded form

Mathematics

2 answers:

aleksklad [387]3 years ago

7 0

Answer:

8 + 0.0 + 0.02+ 0.003

Step-by-step explanation:

Wewaii [24]3 years ago

4 0

Answer:

8+0.02+0.003

hope this helps

have a good day :)

Step-by-step explanation:

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Answer:THE ANSWER IS B MY GUY

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2 years ago

Find the midpoint of the line segment from (1,-1) to (2,4)

Helga [31]

Answer: (1.5, 1.5)

Step-by-step explanation:

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3 years ago

Round to the nearest tenth 6.2x+1.2=8.9

Anuta_ua [19.1K]

Answer:

x = 1.2...

Step-by-step explanation:

Start with the given equation and isolate the variable.

6.2x + 1.2 = 8.9

Subtract 1.2 from each side:

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3 years ago

There is a line that includes the point (2, 10) and has a slope of 1. What is its equation in

rjkz [21]

Answer:

The line is y=1x+8

Step-by-step explanation:

We can use point-slope form of a line to make an equation. Point-slope form is as follows:

y-y₁=m(x-x₁)

**The variables y₁ and x₁ are where we will plug in our given coordinates and the variable m is your slope.

1. Plug in the given info:

y-10=1(x-2)

2. Distribute the 1:

y-10=1x-2

3. Add 10 to both sides:

y=1x+8

5 0

3 years ago

How do you find the length of the curve x=3t−t3x=3t-t^3, y=3t2y=3t^2, where 0≤t≤3√0<=t<=sqrt(3) ?

taurus [48]

It's simple, you just have to do this:

$L=\int\limits_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}~dt$

$x=3t-t^3$

$\frac{dx}{dt}=3-3t^2$

$y=3t^2$

$\frac{dy}{dt}=6t$

replacing

$L=\int\limits_{0}^{\sqrt{3}}\sqrt{\left(3-3t^2\right)^2+\left(6t\right)^2}~dt$

$L=\int\limits_{0}^{\sqrt{3}}\sqrt{9-18t^2+9t^4+36t^2}~dt$

$L=\int\limits_{0}^{\sqrt{3}}\sqrt{9+9t^4+18t^2}~dt$

$L=\int\limits_{0}^{\sqrt{3}}\sqrt{(3t^2+3)^2}~dt$

$L=\int\limits_{0}^{\sqrt{3}}(3t^2+3)~dt$

$L=t^3+3t|_0^{\sqrt{3}}$

$\boxed{\boxed{L=3\sqrt{3}+3\sqrt{3}=6\sqrt{3}}}$

8 0

3 years ago