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Kruka [31]
3 years ago
9

Please help asap will give 20 pts, brain, 5 star and thanks [must include sbs]

Mathematics
2 answers:
stiks02 [169]3 years ago
6 0
61+58=119
180-119=61=angle 2
180-61=119=angle 4
jonny [76]3 years ago
5 0

Step-by-step explanation:

<4 = 61°+58°

= 119° ( exterior angle of a triangle is equal to the sum of two opposite interior angles of a triangle)

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you work out that by v=PI radius squared height divided by 3

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3 years ago
A ladder 3m leans against a wall. The foot of the ladder is 80cm from the wall. The distance the ladder reaches up the wall is:
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Step-by-step explanation:

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3 years ago
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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}

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8 0
1 year ago
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GrogVix [38]

Answer:

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Step-by-step explanation:

3x+2y=3y-2

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x+3x+2=10

4x+2=10

4x=10-2

4x=8

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2+y=10

y=10-2

y=8

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3 0
4 years ago
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hjlf

Answer:

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Step-by-step explanation:

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