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3 years ago

Could you please help me with this problem.

Mathematics

1 answer:

vlabodo [156]3 years ago

6 0

Answer:

please see the attached picture for full solution...

Hope it helps...

Good luck on your assignment....

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DONT SKIP PLZ

irinina [24]

Answer:

The first one: 3.2 units

3 0

3 years ago

Read 2 more answers

Thank you for the help! I would appreciate it if someone could explain this content to me. I need help on both of them. I wish t

maksim [4K]

I'm not 100% sure if I am right, but for number 9 you would pick two points on , let's just say you pick points (0, 1) and (2, -2). Then you would find those points in the table. So in this case the answer would be D. And I'm not really sure how to explain #10 but you basically work out each problem with the graph to see if it makes sense. For example if you try working out problem F it wouldn't make sense since you start of with 10 dollars and earn 2 each day. But on the graph it shows that you are loosing two. Sorry if I didn't explain it right, if you have more questions let me know!

5 0

3 years ago

Can someone please do number 12 and 14 for me ?

KengaRu [80]

12. Yes, because the diagonals of a parallelogram bisect each other.
14. Yes, a parallelogram has opposite sides that are equal.

Hope this helps =)

3 0

3 years ago

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

2 years ago

How do we find the rate of a loan if the total payment is $1,235. The loan was $800 in 5 years

garik1379 [7]

$1235-$800=$435 thus the I which is the simple interest
:I=PRT all over 100
435=800(R) (5)all over 100
43500=4000R
:R=10.88%

7 0

3 years ago