Answer:
so you would rewrite them as (2x-5)(3x-4)
then multiply 2x×3x=6x^2 then -5×-4=20
so the expression is C. 6x^2+20
Answer:
A
Step-by-step explanation:
If two nonadjacent angles formed by two intersecting lines = Vertical angles= congruent
Then
Two non adjacent angles formed by two intersecting lines are congruent
Function y = -2x + 5:
slope of -2
y intercept at 5
x intercept at 2 1/2
function y = x
slope of 1
y intercept at 0
x intercept at 0
A)
to solve a, the following rules are crucial:
(i)
![log_ab-log_ac=log_a \frac{b}{c}](https://tex.z-dn.net/?f=log_ab-log_ac%3Dlog_a%20%5Cfrac%7Bb%7D%7Bc%7D%20)
so the difference of 2 logarithms with the same base, is the logarithm of their division, preserving the same base.
(ii) if
![log_ab=log_ac](https://tex.z-dn.net/?f=log_ab%3Dlog_ac)
then b=c.
so if 2 logarithms with the same base are equal, then the arguments (b and c) are equal as well.
so
![log_3xy-log_3(x-1)=log_36 x^{2} -1](https://tex.z-dn.net/?f=log_3xy-log_3%28x-1%29%3Dlog_36%20x%5E%7B2%7D%20-1)
apply rule (i):
![log_3 \frac{xy}{x-1}=log_3(6 x^{2} -1)](https://tex.z-dn.net/?f=log_3%20%20%5Cfrac%7Bxy%7D%7Bx-1%7D%3Dlog_3%286%20x%5E%7B2%7D%20-1%29)
apply rule (ii):
![\frac{xy}{x-1}=6 x^{2} -1](https://tex.z-dn.net/?f=%20%5Cfrac%7Bxy%7D%7Bx-1%7D%3D6%20x%5E%7B2%7D%20-1)
now 'isolate' y:
![y=(6 x^{2} -1) \frac{(x-1)}{x}](https://tex.z-dn.net/?f=y%3D%286%20x%5E%7B2%7D%20-1%29%20%5Cfrac%7B%28x-1%29%7D%7Bx%7D%20)
b)
some more rules:
(iii)
![log_x a^{n}=nlog_x a](https://tex.z-dn.net/?f=log_x%20a%5E%7Bn%7D%3Dnlog_x%20a)
(iv)
![log_a b= \frac{1}{log_b a}](https://tex.z-dn.net/?f=log_a%20b%3D%20%5Cfrac%7B1%7D%7Blog_b%20a%7D%20)
apply iii and iv:
![log_x25=log_x 5^{2}=2log_x 5=2 \frac{1}{log_5 x}=\frac{2}{log_5 x}](https://tex.z-dn.net/?f=log_x25%3Dlog_x%205%5E%7B2%7D%3D2log_x%205%3D2%20%20%5Cfrac%7B1%7D%7Blog_5%20x%7D%3D%5Cfrac%7B2%7D%7Blog_5%20x%7D)
then substitute
![log_5 x=u](https://tex.z-dn.net/?f=log_5%20x%3Du)
in the equation:
![2u=5- \frac{2}{u}](https://tex.z-dn.net/?f=2u%3D5-%20%5Cfrac%7B2%7D%7Bu%7D%20)
![2u=\frac{5u-2}{u}](https://tex.z-dn.net/?f=2u%3D%5Cfrac%7B5u-2%7D%7Bu%7D%20)
![2u^{2}=5u-2](https://tex.z-dn.net/?f=2u%5E%7B2%7D%3D5u-2%20)
![2u^{2}-5u+2=0](https://tex.z-dn.net/?f=2u%5E%7B2%7D-5u%2B2%3D0)
so now we have a quadratic equation of degree 2,
a=2, b=-5, c=2
the discriminant is
![b^{2}-4ac=25-16=9](https://tex.z-dn.net/?f=%20b%5E%7B2%7D-4ac%3D25-16%3D9)
, the root of it is 3
so the roots are:
![u_1= \frac{-b+3}{2a}= \frac{5+3}{2*2}= \frac{8}{4} =2](https://tex.z-dn.net/?f=u_1%3D%20%5Cfrac%7B-b%2B3%7D%7B2a%7D%3D%20%5Cfrac%7B5%2B3%7D%7B2%2A2%7D%3D%20%5Cfrac%7B8%7D%7B4%7D%20%3D2)
and
![u_2= \frac{-b-3}{2a}= \frac{5-3}{2*2}= \frac{2}{4} = \frac{1}{2}](https://tex.z-dn.net/?f=u_2%3D%20%5Cfrac%7B-b-3%7D%7B2a%7D%3D%20%5Cfrac%7B5-3%7D%7B2%2A2%7D%3D%20%5Cfrac%7B2%7D%7B4%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20)
finally, we convert u's to x's:
![log_5 x=u](https://tex.z-dn.net/?f=log_5%20x%3Du)
means
![x=u^{2}](https://tex.z-dn.net/?f=%20x%3Du%5E%7B2%7D%20)
so for u=2, x=4, and for u=1/2 we have x=1/4
Answers:
A) y=(6 x^{2} -1) \frac{(x-1)}{x}
B) solution set: {4, 1/4}
9514 1404 393
Answer:
x = 20
Step-by-step explanation:
The external angle at V is half the difference of the subtended arcs:
(65° -25°)/2 = x° = 40°/2
x = 20