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shtirl [24]
4 years ago
6

Someone pls help will give brainliest

Mathematics
1 answer:
liberstina [14]4 years ago
5 0
3. 60 divide by 360 times pi times (8)2=33.51 squared inch

4. 120divide by 360 times pi times (5 over 2)2=40.9 squared inch
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If B is the midpoint of AC, and AC= 8x-20, find BC.
Semmy [17]

Answer:

BC = 4x-10

Step-by-step explanation:

\because B is the midpoint of AC.

\therefore BC = \frac{1}{2} AC

\therefore BC = \frac{1}{2} (8x-20)

\therefore BC = \frac{1}{2} \times 2(4x-10)

\huge \purple {\boxed {\therefore BC = 4x-10}}

5 0
3 years ago
What is the correct answer in the photo A. B. C. or .D please explain why
balu736 [363]

Answer:

I think its D)

Step-by-step explanation:

7x−3y+2=0

Add -7x to both sides.

7x−3y+2+−7x=0+−7x

−3y+2=−7x

Add -2 to both sides.

−3y+2+−2=−7x+−2

−3y=−7x−2

Divide both sides by -3.

−3y /−3  =  −7x−2 /−3

which equals d)

I also use this app to convert equations to slope-intercept and it showed me this

8 0
3 years ago
Which could be the missing data item for the given set of data if the median of the complete data set is 15?
Gre4nikov [31]

Answer:

A: 16

Step-by-step explanation:

its in between

5 0
3 years ago
Recall that m(t) = (1/2)^t/h for radioactive decay, where h is the half-life. Suppose that a 500g sample of phosphorus-32 decays
katrin2010 [14]

The question is incomplete, here is the complete question:

Recall that m(t) = m.(1/2)^t/h for radioactive decay, where h is the half-life. Suppose that a 500 g sample of phosphorus-32 decays to 356 g over 7 days. Calculate the half life of the sample.

<u>Answer:</u> The half life of the sample of phosphorus-32 is 14.28days^{-1}

<u>Step-by-step explanation:</u>

The equation used to calculate the half life of the sample is given as:

m(t)=m_o(1/2)^{t/h}

where,

m(t) =  amount of sample after time 't' = 356 g

m_o = initial amount of the sample = 500 g

t = time period = 7 days

h = half life of the sample = ?

Putting values in above equation, we get:

356=500\times (\frac{1}{2})^{7/h}\\\\h=14.28days^{-1}

Hence, the half life of the sample of phosphorus-32 is 14.28days^{-1}

7 0
4 years ago
A surveyor standing on top of a cliff found that the angle of depression of a boat as sea is 22°. if the top of the cliff is 15m
Alex73 [517]

Answer:

From the photo above i interpreted the question and drew a triangle to make it easier to solve

Tan is used to solve the question as we were given opposite (as height of the cliff) and to find Adjacent ( the distance between the foot of the cliff to the boat)

Tan θ =Opp/Adj

Tan 22°=15/a

Cross Multiply

aTanθ=15°

Divide both sides by Tan 22

aTan22/Tan22 = 15/Tan 22

a = 15/Tan 22°

a = 37.13 m

Therefore the distance between the foot of the cliff to the boat is 37.13 metres

5 0
3 years ago
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