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sasho [114]
3 years ago
15

Please help me i am in trouble

Mathematics
2 answers:
motikmotik3 years ago
7 0
I'm not quite sure what the above answer is talking about, but we can work this out by working the problem backwards like so:
175 + 60 = 235
235 x 2 = 470
470 - 40 = 430
430 x 2 = 860
Therefore he had 860 paperclips to start with :))
baherus [9]3 years ago
5 0
1/6 is your answer hope i could help:)

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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
What property is (3x+7)+(2x+8)=3x+7+2x+8
Alex_Xolod [135]

Answer:

distributive property

Step-by-step explanation:

(3x + 7) + (2x + 8)

each of the parenthesis has a multiplier of 1 , placed outside the parenthesis.

then multiplying the terms inside the parenthesis by 1 gives

3x + 7 + 2x + 8

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<    No

Step-by-step explanation:

5 0
3 years ago
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VLD [36.1K]
The answer is : 1/8
I hope it’s help u
4 0
2 years ago
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