Answer:
×3+![2^{3}](https://tex.z-dn.net/?f=2%5E%7B3%7D)
Step-by-step explanation:
Firstly, two cubed will be
.
(Remember cubed has a exponent of 3, but squared has a exponent of 2.)
"Three times the square root of 100" is 3×![\sqrt{100}](https://tex.z-dn.net/?f=%5Csqrt%7B100%7D)
Then, there is "More than"
Which flips the whole expression.
So instead of ![2^{3} +](https://tex.z-dn.net/?f=2%5E%7B3%7D%20%2B)
×![\sqrt{100}](https://tex.z-dn.net/?f=%5Csqrt%7B100%7D)
It will be
×
+![2^{3}](https://tex.z-dn.net/?f=2%5E%7B3%7D)
Answer:
85 quarters and 135 dimes
Step-by-step explanation:
its basically a guess and check
formula - 0.25q + 0.10d = $34.75
0.25 x 85 + 0.10 x 135 = 34.75
9 x 3.5 = 31.5
2 x 2 = 4
2 x 2 = 4 / 2 = 2
2 x 5 = 10/ 2 = 5
31.5 + 4 + 2 + 5 = 42.5^2
First we'll do two basic steps. Step 1 is to subtract 18 from both sides. After that, divide both sides by 2 to get x^2 all by itself. Let's do those two steps now
2x^2+18 = 10
2x^2+18-18 = 10-18 <<--- step 1
2x^2 = -8
(2x^2)/2 = -8/2 <<--- step 2
x^2 = -4
At this point, it should be fairly clear there are no solutions. How can we tell? By remembering that x^2 is never negative as long as x is real.
Using the rule that negative times negative is a positive value, it is impossible to square a real numbered value and get a negative result.
For example
2^2 = 2*2 = 4
8^2 = 8*8 = 64
(-10)^2 = (-10)*(-10) = 100
(-14)^2 = (-14)*(-14) = 196
No matter what value we pick, the result is positive. The only exception is that 0^2 = 0 is neither positive nor negative.
So x^2 = -4 has no real solutions. Taking the square root of both sides leads to
x^2 = -4
sqrt(x^2) = sqrt(-4)
|x| = sqrt(4)*sqrt(-1)
|x| = 2*i
x = 2i or x = -2i
which are complex non-real values
15 is the answer of this question