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castortr0y [4]
3 years ago
7

Rearrange the formula to calculate the diameter of a circle.

Mathematics
2 answers:
julia-pushkina [17]3 years ago
7 0

a=1/4pid^2
4a=pid^2
4a/pi=d^2
square both sides to get d
Answer: Square root of 4a over pi equals d.
Andrew [12]3 years ago
5 0
To find your answer, you want to get everything but the variable "d," the diameter, onto one side of the equal sign, leaving d by itself.

1) Start by multiplying both sides by 4 to get rid of the 4 in the denominator on the right. bringing it to the other side:
A =  \frac{1}{4}  \pi  d^{2}  \\
4(A) =  4(\frac{1}{4}  \pi  d^{2} )\\
4A =  \pi  d^{2}

2) Divide both sides by π to move the <span>π</span> to the other side of the equation (I put the d squared on the left, but you can keep it on the right):
4A =  \pi  d^{2}  \\&#10; d^{2}  = \frac{4A}{ \pi }

3) Finally, take the square root of both sides to get the value of d:
d^{2} = \frac{4A}{ \pi } \\&#10;d =  \sqrt{\frac{4A}{ \pi } }

-----

Answer: d = \sqrt{\frac{4A}{ \pi } }


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Rewrite in mathematical language. Every negative integer J is less than or equal to its inverse.
Luda [366]

Answer:

-J \leq J

Step-by-step explanation:

Given

Negative integer J

Required

Represent as an inequality of its inverse

The question didn't state if it's additive inverse or multiplicative inverse;

<em>Since the question has to do with negation, I'll assume it's an additive inverse</em>

<em></em>

The inverse of -J is +J

To represent as an inequality (less than or equal), we have:

-J \leq +J

Solving further, it gives

-J \leq J

3 0
3 years ago
Two integers, n and p, have a product of -18. What is the largest possible sum of n and p?
Naily [24]

Answer:

17

Step-by-step explanation:

The two integers could be -1 and 18.

18 + -1 = 17

Any other combination of integers that have a product of -18 would have a smaller sum.

4 0
3 years ago
A hungry elf ate 30 of your muffins. that was 5/8 of all of them! how many are left?
Lilit [14]
There18 muffins left
4 0
2 years ago
The measure of an angle in standard position is given. Find two positive angles and two negative angles that are coterminal with
Pani-rosa [81]
Notice the picture below

negative angles, are just angles that go "clockwise", namely, the same direction a clock hands move hmmm so....  and one revolution is just 2π

now, you can have angles bigger than 2π of course, by simply keep going around, so, if you go around 3 times on the circle, say "counter-clockwise", or from right-to-left, counter as a clock goes, 3 times or 3 revolutions will give you an angle of 6π, because 2π+2π+2π is 6π

now... say... you have this angle here... let us find another that lands on that same spot

by simply just add 2π to it :)  

\bf \cfrac{7}{6}+2=\cfrac{19}{6}\qquad thus\qquad \cfrac{7\pi} {6}+2\pi =\cfrac{19\pi }{6}&#10;\\\\\\&#10;\cfrac{19\pi }{6}\impliedby co-terminal\ angle

now, that's a positive one
and  \bf \cfrac{7}{6}-2=-\cfrac{5}{6}\qquad thus\qquad \cfrac{7\pi} {6}-2\pi =-\cfrac{5\pi }{6}&#10;\\\\\\&#10;-\cfrac{5\pi }{6}\impliedby \textit{is also a co-terminal angle}

to get more, just keep on subtracting or adding 2π


8 0
3 years ago
Can CD be changed to DC
vichka [17]
I would say no thats my anwser
8 0
3 years ago
Read 2 more answers
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