Answer:
I think it's d. C and D
Step-by-step explanation:
it needs to be 20 characters long so don't mind this part
If 2^y = 3, then:
(log to the base 2 of) 2^y = (log to the base 2 of) 3
or
y = log to the base 2 of 3
Note that y turns out to be a constant here.
We need to identify the graph that depicts (log to the base 2) of x, and once we have that, draw a vertical line thru x = 3 to estimate y.
Only the 3rd graph qualifies here. Note that all basic log graphs have an x-intercept of (1,0), as the 3rd graph does.
If we draw on the 3rd graph a vertical line thru x=3, we see that this vert. line intersects the curve at roughly y=1.5.
A.
the radicand is b²-4ac
if it is in form ax²+bx+c
when it is
greater than 0, then 2 real roots
equal to 0, then 1 real root
less than 0, then no real roots
so
a=4
b=-12
c=10
b²-4ac=(-12)²-4(4)(10)=144-160<0
so 0 real roots
B.
using quadratic formula because it is easier
for
ax²+bx+c=0
![x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%2B%2F-%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
so
for 2x²-13x+21
a=2
b=-13
c=21
![x=\frac{-(-13)+/-\sqrt{(-13)^2-4(2)(21)}}{2(2)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-%28-13%29%2B%2F-%5Csqrt%7B%28-13%29%5E2-4%282%29%2821%29%7D%7D%7B2%282%29%7D)
![x=\frac{13+/-\sqrt{169-168}}{4}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B13%2B%2F-%5Csqrt%7B169-168%7D%7D%7B4%7D)
![x=\frac{13+/-\sqrt{1}}{4}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B13%2B%2F-%5Csqrt%7B1%7D%7D%7B4%7D)
![x=\frac{13+/-1}{4}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B13%2B%2F-1%7D%7B4%7D)
x=(13+1)/4 or (13-1)/4
x=14/4 or 12/4
x=7/2 or 3
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w = 3(2a + b) - 4
First, add 4 to both sides:
w (+4) = 3(2a + b) - 4 (+4)
w + 4 = 3(2a + b)
Next, divide 3 from both sides:
(w + 4)/3 = (3(2a + b))/3
(w + 4)/3 = 2a + b
Next, subtract b from both sides:
((w + 4)/3) - b = 2a + b (-b)
(w + 4)/3 - b = 2a
Finally, divide 2 from both sides:
((w + 4)/3 - b)/2 = (2a)/2
a = (((w + 4)/3) - b)/2
a = (((w + 4)/3) - b)/2 is your answer.
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