Answer:
A perfect square is a whole number that is the square of another whole number.
n*n = N
where n and N are whole numbers.
Now, "a perfect square ends with the same two digits".
This can be really trivial.
For example, if we take the number 10, and we square it, we will have:
10*10 = 100
The last two digits of 100 are zeros, so it ends with the same two digits.
Now, if now we take:
100*100 = 10,000
10,000 is also a perfect square, and the two last digits are zeros again.
So we can see a pattern here, we can go forever with this:
1,000^2 = 1,000,000
10,000^2 = 100,000,000
etc...
So we can find infinite perfect squares that end with the same two digits.
<span>since sin and cos = each other at pi/4; take your integrals from 0 to pi/4
</span><span>[S] cos(t) dt - [S] sin(t) dt ;[0,pi/4]
</span>
<span>to revolve it around the x axis;
we do a sum of areas
[S] 2pi [f(x)]^2 dx
</span>
<span>take the cos first and subtract out the sin next; like cutting a hole out of a donuts.
</span><span>pi [S] cos(x)^2 dx - [S] sin(x)^2 dx ; [0,pi/4]
</span>
<span>cos(2t) = 2cos^2 - 1
cos^2 = (1+cos(2t))/2
</span>
<span>1/sqrt(2) - (-1/sqrt(2) +1)
1/sqrt(2) + 1/sqrt(2) -1
(2sqrt(2) - sqrt(2))/sqrt(2) = sqrt(2)/sqrt(2) = 1</span>
In this question, you're solving for d.
Solve for d:
5(-6 - 3d) = 3(8+7d)
Use the distributive property:
-30 - 15d = 3(8+7d)
-30 - 15d = 24 + 21d
Add 30 to both sides:
-15d = 54 + 21d
Subtract 21d from both sides
-36d = 54
Divide both sides by -36
d = -3/2
Answer:
d = -3/2 or -1.5
Answer:
16 and 2
Step-by-step explanation:
16+2=18
16-2=14
