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larisa [96]
3 years ago
9

Work out the circumference of this circle.

Mathematics
1 answer:
tigry1 [53]3 years ago
3 0
C=pid
Since d=9m and pi=3.142, C=3.142*9m=28.278
Rounded to 1 decimal point =28.3
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Please help me! I'll give brainliest!
german
Tan A  = BC/AB = 15/8

answer is B. 15/8
6 0
3 years ago
Read 2 more answers
If n+h/5=f+9/9, then n/5=
Ann [662]
------------------------------------------------------------------
Question
------------------------------------------------------------------
\boxed { \frac{n+h}{5}  =  \frac{f+9}{9}}

------------------------------------------------------------------
Split the fraction on the left
------------------------------------------------------------------
\boxed { \frac{n}{5} + \frac{h}{5}  = \frac{f + 9}{9}}

------------------------------------------------------------------
Take away h/5 from both sides
------------------------------------------------------------------
\boxed { \frac{n}{5}  = \frac{f+9}{9} - \frac{h}{5}}

------------------------------------------------------------------
Change the denominator to be the same
------------------------------------------------------------------
\boxed { \frac{n}{5} = \frac{5f+45}{45} - \frac{9h}{45}}

------------------------------------------------------------------
Put it into single fraction
------------------------------------------------------------------
\boxed { \frac{n}{5} = \frac{5f+45-9h}{45} }

-------------------------------------------------------------------
Rearrange (This step may not be necessary)
------------------------------------------------------------------
\boxed {\frac{n}{5} = \frac{5f-9h+ 45}{45} }


\bf \Longrightarrow \ Answer \ : \boxed {\boxed {\frac{n}{5} = \frac{5f-9h+ 45}{45} }}

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Write a problem that can be solved using a flowchart and working backward. Then draw the flowchart and solve the problem.
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What is the constants and coefficients for 2m+3m-m
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Suppose a spherical balloon grows in such a way that after t seconds, its volume is V = 4 sqrt(t) cm3. What is the volume of the
Arisa [49]
:<span>  </span><span>You need to know the derivative of the sqrt function. Remember that sqrt(x) = x^(1/2), and that (d x^a)/(dx) = a x^(a-1). So (d sqrt(x))/(dx) = (d x^(1/2))/(dx) = (1/2) x^((1/2)-1) = (1/2) x^(-1/2) = 1/(2 x^(1/2)) = 1/(2 sqrt(x)). 

There is a subtle shift in meaning in the use of t. If you say "after t seconds", t is a dimensionless quantity, such as 169. Also in the formula V = 4 sqrt(t) cm3, t is apparently dimensionless. But if you say "t = 169 seconds", t has dimension time, measured in the unit of seconds, and also expressing speed of change of V as (dV)/(dt) presupposes that t has dimension time. But you can't mix formulas in which t is dimensionless with formulas in which t is dimensioned. 

Below I treat t as being dimensionless. So where t is supposed to stand for time I write "t seconds" instead of just "t".

Then (dV)/(d(t seconds)) = (d 4 sqrt(t))/(dt) cm3/s = 4 (d sqrt(t))/(dt) cm3/s = 4 / (2 sqrt(t)) cm3/s = 2 / (sqrt(t)) cm3/s. 

Plugging in t = 169 gives 2/13 cm3/s.</span>
4 0
3 years ago
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