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Margaret [11]
3 years ago
5

Kale is a leafy green vegetable that is low in calories

Mathematics
2 answers:
Ivahew [28]3 years ago
7 0

Answer:

Yes

Step-by-step explanation:

Its 100% true :)

disa [49]3 years ago
5 0

Answer:

yes this is true

Step-by-step explanation:

HAVE A GREAT DAY!!!!!!!!!!!!!!

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Please help i’ll give u a brainliest
lubasha [3.4K]

Answer:

first sphere

volume = 113.142 m^3

2nd sphere

radius= 6 m

first cone

volume = 314.285 m^3

2nd cone

height=8 m

1st cylinder

volume=88 m^3

2nd cylinder

height=3m

7 0
3 years ago
10y-50=<br> 10y−50=<br> \,\,-20<br> −20
densk [106]

Answer:

Therefore the value of 'y' is,

y=3

Step-by-step explanation:

Given:

10y-50=-20

To Find:

y= ?

Solution:

10y-50=-20    ........Given

Step 1. Adding 50 to both the side we get

10y-50+50=-20+50\\10y=30

Step 2. Dividing by 10 on both the side we get

\dfrac{10y}{10}=\dfrac{30}{10}\\\\y=3

Therefore the value of 'y' is,

y=3

6 0
3 years ago
Plzzzzzzzzzzzzzzzzzzzzzzzzzzzz help me plzzzzzzzzz this due right now
marusya05 [52]

Answer:

i will answer it if u raise the point score lol i just need a friend

3 0
3 years ago
What is the exact volume of the cone?
AlekseyPX
The volume is 508.94 cm^3

7 0
3 years ago
Read 2 more answers
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
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