Answer:
1. option B is correct i.e., 
2. option D is correct i.e.,
.
3. option B is correct i.e.,
.
Step-by-step explanation:
1. Since the given equation is:

As we know that 
So, the given equation can be represented as:

2. Given equation is:

As we know that 
so the given equation can be represented as:
.
3. Given equation is:

by using the properties
,
and 

= 
= 
= 
= 
=
.
which is option B.
Answer:
x=120
Step-by-step explanation:
Answer:
The third choice is the one you want
Step-by-step explanation:
If we are to write the equation of a line perpendicular to WX, we first must determine what the slope of the WX is, because the line perpendicular to WX has a slope that is the flip of the slope of WX with the opposite sign. Solving for y takes care of finding the slope of WX:
2x + y = -5 so
y = -2x - 5
The slope is -2. That means that the reciprocal slope is 1/2. Using that slope along with the coordinates x = -1 and y = -2, we first write the line using point-slope form and then solve it for y. Start by filling in the m, the x value and the y value:

Getting rid of the double negatives gives us:

Distributing then gives us:

And finally solving for y (I am going to express the 2 on the left as 4/2 when I move it by subtraction in order to add those fractions):

And the final equation in slope-intercept form is:

Step-by-step explanation:
z= 21/12
divide bothside by 12
shekinan
Answer:
b. (1, 3, -2)
Step-by-step explanation:
A graphing calculator or scientific calculator can solve this system of equations for you, or you can use any of the usual methods: elimination, substitution, matrix methods, Cramer's rule.
It can also work well to try the offered choices in the given equations. Sometimes, it can work best to choose an equation other than the first one for this. The last equation here seems a good one for eliminating bad answers:
a: -1 -5(1) +2(-4) = -14 ≠ -18
b: 1 -5(3) +2(-2) = -18 . . . . potential choice
c: 3 -5(8) +2(1) = -35 ≠ -18
d: 2 -5(-3) +2(0) = 17 ≠ -18
This shows choice B as the only viable option. Further checking can be done to make sure that solution works in the other equations:
2(1) +(3) -3(-2) = 11 . . . . choice B works in equation 1
-(1) +2(3) +4(-2) = -3 . . . choice B works in equation 2