The line y = kx + 4, where k is a constant, is

Mathematics

1 answer:

tresset_1 [31]3 years ago

4 0

Answer:

(d - 4) / c

Step-by-step explanation:

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Answer:

35 + 80 = 115

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How to solve for -3x squared + 2y squared + 5xy - 2y + 5x squared - 3y squared when x = 0.5 and y = -1/10??

Pani-rosa [81]

Answer:

0.44

Step-by-step explanation:

-3x^2 + 2y^2 + 5xy - 2y +5x^2 - 3y^2

Combine like terms

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3 years ago

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Dennis_Churaev [7]

8x+Y=-16
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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.