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2 years ago

i’ll mark brainliest Can someone help me answer #23 A and B. Thanks in advance.

Mathematics

1 answer:

Mila [183]2 years ago

3 0

I would set D = 816000 and then solve

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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

Nicole earns $40,768 per year, what are her weekly gross earnings?

Karolina [17]

($40768/y)(y/52w)=$784/week

Seven hundred and eighty-four dollars per week.

4 0

2 years ago

Read 2 more answers

Find the common ratio of the geometric sequence 11, -55, 275,...11,−55,275,...

Kaylis [27]

Answer:

r = -5

Step-by-step explanation:

common ratio (r) is found be dividing one term by its preceding term:

r = 275 ÷ -55 = -5

or, r = -55 ÷ 11 = -5

Therefore, r = -5

3 0

2 years ago

-3 ( -4 - 5x) = -18 i'll help with your questions :)))

maria [59]

Answer:

-15

Step-by-step explanation:

-3(-4-5x)= -18

12+15x= -18-12

15x= -30

x= -30÷15

x= -15

6 0

3 years ago

I need help with this please its due today :(

yKpoI14uk [10]

Answer:

$a) \ 6^2 \times 6^3 = 6^{2+3} = 6^5 = 7776\\\\b)\ \frac{(-3)^5 }{(-3)^2} = (-3)^{5-2} = (-3)^3 = -27\\\\c)\ [(-2)^3]^4 = (-2)^{3\times 4} = (-2)^{12} = 4096 \\\\d)\ (\frac{1}{999})^0 = 1, anything \ raise \ to \ zero = 1\\\\e) \ (\frac{4}{3})^{-2} = 4^{-2} \times 3^2 = \frac{9}{16}\\\\f) \ \frac{2^5 \times 2^9}{2^{16}} = 2^{5+9-16} = 2^{-2} = \frac{1}{4}$

4 0

2 years ago

*i’ll mark brainliest* Can someone help me answer #23 A and B. Thanks in advance.

i’ll mark brainliest Can someone help me answer #23 A and B. Thanks in advance.