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Artist 52 [7]
3 years ago
12

Given the following formula, solve for v. s = 1/2a^2v + c

Mathematics
2 answers:
dexar [7]3 years ago
8 0

Answer:

c

Step-by-step explanation:

correct

weeeeeb [17]3 years ago
7 0

Since there is no initial condition, the exact value of v cannot be determined, but you can set the equation up for a general evaluation of the situation.

s = 1/2a^2*v+c

First you need to subtract c from both sides to get

s-c = 1/2a^2*v

then you can just divide both sides by 1/2a^2 to get v

(2(s-c))/a^2=v

when dividing a fraction, such as 1/2, make sure to keep in mind that you're really multiplying the flipped version, so that dividing by 1/2 means multiplying by 2.

You might be interested in
Evaluate the expression 11C4.
kicyunya [14]

The correct answer is: 330

Explanation:

The formula for combination is given as follows:

^nC_r = \frac{n!}{r! (n-r)!} --- (1)

Where n = 11

r = 4

Plug in the values in equation (1) and calculate:

^nC_r = \frac{n!}{r! (n-r)!} \\ ^{11}C_{4} = \frac{11!}{4! (11-4)!} \\ ^{11}C_{4} = \frac{11*10*9*8*(7!)}{4! (7)!} \\ ^{11}C_{4} = \frac{11*10*9*8}{4!} \\ ^{11}C_{4} = \frac{11*10*9*8}{4*3*2*1} \\ ^{11}C_{4} = 330 \\

Hence the correct answer is 330.

8 0
3 years ago
Read 2 more answers
WILL GIVE BRAINLIEST IF YOU HELP MEE
yarga [219]

Answer: A: Whole Number B:Integer and C: Rational are all correct.

For the question: In the figure, AB is divided into equal parts. The coordinates of point A are (2, 4), and the coordinates of point B are (10, 6). Match each pair of coordinates to the corresponding point on AB.

Answer:The coordinates of the point are (6 , 5).

For the question: Imani and Abedi drive to work. Imani drives 66 miles in 1.5 hours. Abedi drives 56 km in 1 hour 15 min. Work out the difference between their average speeds in km/h. 1 mile = 1.6 km

Answer:25.6 km/h

I couldn't help you on all of them, but I helped you on some. Hope this helps :D

8 0
1 year ago
<img src="https://tex.z-dn.net/?f=3a-4%5Cleq5" id="TexFormula1" title="3a-4\leq5" alt="3a-4\leq5" align="absmiddle" class="latex
Gnesinka [82]
It depends, what does a mean? If you know just add what a means to 3 then subtract 4.
4 0
3 years ago
(05.03)An irregular polygon is shown below:
SVEN [57.7K]
9units sorry for not showing work though but glad I can help
7 0
3 years ago
I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).
Aleksandr-060686 [28]

Answer:

\frac{7\pi}{24} and \frac{31\pi}{24}

Step-by-step explanation:

\sqrt{3} \tan(x-\frac{\pi}{8})-1=0

Let's first isolate the trig function.

Add 1 one on both sides:

\sqrt{3} \tan(x-\frac{\pi}{8})=1

Divide both sides by \sqrt{3}:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

Now recall \tan(u)=\frac{\sin(u)}{\cos(u)}.

\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}

or

\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}

The first ratio I have can be found using \frac{\pi}{6} in the first rotation of the unit circle.

The second ratio I have can be found using \frac{7\pi}{6} you can see this is on the same line as the \frac{\pi}{6} so you could write \frac{7\pi}{6} as \frac{\pi}{6}+\pi.

So this means the following:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

is true when x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

where n is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

Add \frac{\pi}{8} on both sides:

x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi

Find common denominator between the first two terms on the right.

That is 24.

x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi

x=\frac{7\pi}{24}+n \pi (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval [0,2\pi).

So if \sqrt{3} \tan(x-\frac{\pi}{8})-1=0 and we let u=x-\frac{\pi}{8}, then solving for x gives us:

u+\frac{\pi}{8}=x ( I just added \frac{\pi}{8} on both sides.)

So recall 0\le x.

Then 0 \le u+\frac{\pi}{8}.

Subtract \frac{\pi}{8} on both sides:

-\frac{\pi}{8}\le u

Simplify:

-\frac{\pi}{8}\le u

-\frac{\pi}{8}\le u

So we want to find solutions to:

\tan(u)=\frac{1}{\sqrt{3}} with the condition:

-\frac{\pi}{8}\le u

That's just at \frac{\pi}{6} and \frac{7\pi}{6}

So now adding \frac{\pi}{8} to both gives us the solutions to:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}} in the interval:

0\le x.

The solutions we are looking for are:

\frac{\pi}{6}+\frac{\pi}{8} and \frac{7\pi}{6}+\frac{\pi}{8}

Let's simplifying:

(\frac{1}{6}+\frac{1}{8})\pi and (\frac{7}{6}+\frac{1}{8})\pi

\frac{7}{24}\pi and \frac{31}{24}\pi

\frac{7\pi}{24} and \frac{31\pi}{24}

5 0
3 years ago
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