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dangina [55]
3 years ago
6

4.1 * 10 to the 2nd power in standard form

Mathematics
2 answers:
ivolga24 [154]3 years ago
6 0
The answer would be 410 hope this helps
marshall27 [118]3 years ago
3 0
The answer is 410 but i cant put in standard form now im in hurry

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Given the function f(x)=3x -2 and g(x)= x+2/3 complete parts A and B.
deff fn [24]

First, I should point out that g(x) should be written as g(x)=(x+2)/3, otherwise the problem is confusing.  

f(x)=3x-2 \enspace g(x)=\frac{x+2}{3}

(A) f(g(x))=3(\frac{x+2}{3})-2=x\\g(f(x))=\frac{3x-2+2}{3}=x

(B) Since f(g(x))=x and g(f(x))=x, it holds that

f(g(x))=g(f(x)) for all x. This means the composed functions are *identical*

7 0
4 years ago
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I need help please! I really do! Thank you.
SCORPION-xisa [38]
\bf \qquad \textit{parabola vertex form}\\\\

\begin{array}{llll}
{y=(x-{{ h}})^2+{{ k}}}\\\\
\end{array} \qquad\qquad  vertex\ ({{ h}},{{ k}})

notice your vertex there, 1, -3
h = 1
k = -3
7 0
3 years ago
Given these angles: m∠3 = 82°, m∠4 = 98°.
RSB [31]
These angles are supplementary.
Complementary angles add up to 90 degrees. Supplementary angles add up to 180 degrees. 82 degrees + 98 degrees = 180 degrees.
8 0
3 years ago
Subtract four-fifths from 8 times b
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What does b = 4/5 8b?
4 0
4 years ago
A container manufacturing company was contracted to design and manufacture cylindrical cans for fruit juice. The volume of each
gtnhenbr [62]

The volume of the cylinder is the amount of fruit juice it can contain.

The relationship between the volume and the surface area is:

\mathbf{A = \pi (\sqrt[3]{\frac{V}{2\pi}})^2 + \frac{0.946}{(\sqrt[3]{\frac{V}{2\pi}})}}

The given parameter is:

\mathbf{V = 0.946}

The volume of a cylinder is calculated as:

\mathbf{V = \pi r^2 h}

Make h the subject

\mathbf{h = \frac{V}{ \pi r^2}}

The surface area (A) of a cylinder is:

\mathbf{A = \pi r^2 + \pi rh}

Substitute \mathbf{h = \frac{V}{ \pi r^2}}

\mathbf{A = \pi r^2 + \pi r \times \frac{V}{\pi r^2}}

\mathbf{A = \pi r^2 + \frac{V}{r}}

Differentiate

\mathbf{A' = 2\pi r - Vr^{-2}}

Set to 0

\mathbf{2\pi r - Vr^{-2} = 0}

Rewrite as:

\mathbf{ 2\pi r = Vr^{-2}}

Multiply through by r^2

\mathbf{ 2\pi r^3 = V}

Solve for r

\mathbf{r = \sqrt[3]{\frac{V}{2\pi}}}

\mathbf{A = \pi r^2 + \frac{0.946}{r}}

So, we have:

\mathbf{A = \pi (\sqrt[3]{\frac{V}{2\pi}})^2 + \frac{0.946}{(\sqrt[3]{\frac{V}{2\pi}})}}

Hence, the relationship between the volume and the surface area is:

\mathbf{A = \pi (\sqrt[3]{\frac{V}{2\pi}})^2 + \frac{0.946}{(\sqrt[3]{\frac{V}{2\pi}})}}

Read more about surface areas and volumes at:

brainly.com/question/3628550

8 0
2 years ago
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