Answer:
5 solutions.
x=0
x=0
x=0
x=1
x=-1
Step-by-step explanation:
To solve this, we need to handle the radical function somehow, an easy approach is to raise to square power the whole equation, this way you can get rid of it.
sqr( x^(3/2) = x^(5/2) )=
x^3 = x^5
Now, lets group terms
x^5-x^3==0
GCF x^3
x^3(x^2-1)=0
The above equation gives us already 5 answer
In order to satisfy the equation there are two choices:
x^3=0 (a)
or (x^2-1)=0 (b)
From (a), we have that x=0, thus given one solution from here.
However, the amount of solutions is not just one, its 3, because the amount of solutions is linked to the exponent of the variable.
the other equation (b)
gives us x =sqrt(1), leading to x= -1 or x=+1
We have two solutions here, again, the exponent of the variable already tells us the amount of solutions,
Alright, let's see.
#1. The answer is "more".
#2. "A redder"
#3. "A bluer"
The last two answers are theoretical, but I believe correct.
Answer:
18x
Step-by-step explanation:
2(3x+6x)
Let's test it out.
Our first pentagonal prism will have a base edge length of 3 in and a height of 3 in. One formula I found for the surface area of a pentagonal prism is
![SA = 5ah+ \frac{1}{2} \sqrt{5(5+2 \sqrt{5}) } *a^{2}](https://tex.z-dn.net/?f=SA%20%3D%205ah%2B%20%5Cfrac%7B1%7D%7B2%7D%20%5Csqrt%7B5%285%2B2%20%5Csqrt%7B5%7D%29%20%7D%20%2Aa%5E%7B2%7D%20)
, where <em>a</em> is the base edge length and <em>h</em> is the height. I was able to simplify the term being multiplied to <em>a</em><em>² </em>like this:
![\frac{1}{2} \sqrt{5(5+2 \sqrt{5}) } = \\ \\ \frac{1}{2} \sqrt{25+10 \sqrt{5}} = \\ \\ \frac{1}{2} \sqrt{25+22.360679775} = \\ \\ \frac{1}{2} \sqrt{47.360679775}= \\ \\ \frac{1}{2} (6.88190960236)= \\ \\ 3.44095480118](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5Csqrt%7B5%285%2B2%20%5Csqrt%7B5%7D%29%20%7D%20%3D%20%5C%5C%20%5C%5C%20%5Cfrac%7B1%7D%7B2%7D%20%5Csqrt%7B25%2B10%20%5Csqrt%7B5%7D%7D%20%3D%20%5C%5C%20%5C%5C%20%5Cfrac%7B1%7D%7B2%7D%20%5Csqrt%7B25%2B22.360679775%7D%20%3D%20%5C%5C%20%5C%5C%20%5Cfrac%7B1%7D%7B2%7D%20%5Csqrt%7B47.360679775%7D%3D%20%5C%5C%20%5C%5C%20%5Cfrac%7B1%7D%7B2%7D%20%286.88190960236%29%3D%20%5C%5C%20%5C%5C%203.44095480118)
The formula looks like this now:
![SA = 5ah+3.44095480118a^{2}](https://tex.z-dn.net/?f=SA%20%3D%205ah%2B3.44095480118a%5E%7B2%7D)
. We can plug in our values for <em>a </em>and <em>h </em>to find the surface area of the prism:
![SA = 5(3)(3)+3.44095480118*3^{2} \\ SA =45+3.44095480118*9 \\ SA =45+30.9685932106 \\ SA =75.9685932106](https://tex.z-dn.net/?f=SA%20%3D%205%283%29%283%29%2B3.44095480118%2A3%5E%7B2%7D%20%5C%5C%20SA%20%3D45%2B3.44095480118%2A9%20%5C%5C%20SA%20%3D45%2B30.9685932106%20%5C%5C%20SA%20%3D75.9685932106)
The surface area of our initial prism is approximately 76 in². Let's quadruple the dimensions (<em>a</em> is 12 and <em /><em>h </em>is 12) and plug them into the formula:
![SA = 5(12)(12)+3.44095480118*12^{2} \\ SA = 720+3.44095480118*144 \\ SA = 720+495.49749137 \\ SA = 1215.49749137](https://tex.z-dn.net/?f=SA%20%3D%205%2812%29%2812%29%2B3.44095480118%2A12%5E%7B2%7D%20%5C%5C%20SA%20%3D%20720%2B3.44095480118%2A144%20%5C%5C%20SA%20%3D%20720%2B495.49749137%20%5C%5C%20SA%20%3D%201215.49749137)
The surface area of the new prism is approximately 1215 in². To finally answer this question, let's divide the second prism's surface area by the first's to see if we get 8. The first prism's surface area should fit into the second's about 8 times if the statement is true:
1215 ÷ 76 ≈ 16
The statement is incorrect. If the dimensions of a pentagonal prism are quadrupled, then the surface area of the prism is multiplied by ~16, not 8.
Answer:
D
Step-by-step explanation:
![2x^2+16x-8=0](https://tex.z-dn.net/?f=2x%5E2%2B16x-8%3D0)
Simplify:
![x^2+8x=4](https://tex.z-dn.net/?f=x%5E2%2B8x%3D4)
Rewrite in the form
:
![x^2+8x=4](https://tex.z-dn.net/?f=x%5E2%2B8x%3D4)
Add
to both sides:
![x^2+8x+4^2=4+4^2](https://tex.z-dn.net/?f=x%5E2%2B8x%2B4%5E2%3D4%2B4%5E2)
![x^2+8x+4^2=20](https://tex.z-dn.net/?f=x%5E2%2B8x%2B4%5E2%3D20)
![\left(x+4\right)^2=20](https://tex.z-dn.net/?f=%5Cleft%28x%2B4%5Cright%29%5E2%3D20)
or ![x+4=-\sqrt{20}](https://tex.z-dn.net/?f=x%2B4%3D-%5Csqrt%7B20%7D)
x = -4±![\sqrt{20}](https://tex.z-dn.net/?f=%5Csqrt%7B20%7D)
x= -4±![2\sqrt{5}](https://tex.z-dn.net/?f=2%5Csqrt%7B5%7D)