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Ede4ka [16]
3 years ago
14

Write the number 208000478 in expanded form

Mathematics
2 answers:
Alja [10]3 years ago
5 0
200000000+80000000+400+70+8
Lisa [10]3 years ago
3 0
20,000,000+80,000+4,000+78

It's been a while since i learned this..

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2. Which is equivalent to the calculator output 8.954 E - 15?
anyanavicka [17]

Answer: 8.954 x 10^-15

Step-by-step explanation:

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2 years ago
Ms.Liang is building a deck that is 2 2/9 yards long and 3 2/5 yards wide what is the area of her deck.
choli [55]

Answer:

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breadth = 17/5 yards

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Questions attached as screenshot below:Please help me I need good explanations before final testI pay attention
Nikitich [7]

The acceleration of the particle is given by the formula mentioned below:

a=\frac{d^2s}{dt^2}

Differentiate the position vector with respect to t.

\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}

Differentiate both sides of the obtained equation with respect to t.

\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}

Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.

\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}

The initial position is obtained at t=0. Substitute t=0 in the given position function.

\begin{gathered} s(0)=-23\times0+65 \\ =65 \end{gathered}

8 0
1 year ago
Andre says I'm multiplied 4 by 5 then cubed the results select all expressions are equal Andres answer a 4×5 to the power of 3B4
Charra [1.4K]

Answer:

(4 * 5)^3 ;

8000

Step-by-step explanation:

Andre's statement :

Multiply 4 by 5 = 4 * 5 = 20

Cube the result = 20³

The equal expression from the option :

A.) (4 * 5)^3

Also, (4 * 5)^3 = 20^3 = 20 * 20 *20 = 8000

5 0
2 years ago
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