Answer:
A is correct
Step-by-step explanation:
A has no solution because since the slopes are the same, setting both equations equal to each other will cancel out the slopes and result in 8=16, meaning no solution will make both sides equal to each other.
B does have a solution because of different slopes
C does have a solution because of different slopes
D does have a solution because of different slopes
The equation of a line that is perpendicular to the given line is y = –4x – 16.
Solution:
The equation of a line given is y = 0.25x – 7
Slope of the given line(
) = 0.25
Let
be the slope of the perpendicular line.
Passes through the point (–6, 8).
<em>If two lines are perpendicular then the product of the slopes equal to –1.</em>




Point-slope intercept formula:

and 
Substitute these in the formula, we get



Add 8 on both sides of the equation.


Hence the equation of a line that is perpendicular to the given line is
y = –4x – 16
Answer:
a. 
b. 
c. 
d. -6t - 5
Step-by-step explanation:
When a number has the same variable as another, you can add or subtract them
So for the first one, you can do

(But don't forget the
!)
So it is:

Continue this process for the rest of them
Given

We have to set the restraint

because a square root is non-negative, and thus it can't equal a negative number. With this in mind, we can square both sides:

The solutions to this equation are 7 and -2. Recalling that we can only accept solutions greater than or equal to -1, 7 is a feasible solution, while -2 is extraneous.
Similarly, we have

So, we have to impose

Squaring both sides, we have

The solutions to this equation are 5 and 10. Since we only accept solutions greater than or equal to 7, 10 is a feasible solution, while 5 is extraneous.
Answer:
Yes
Step-by-step explanation:
This is true because it is asking you if this is less than or EQUAL TO. 4 = 4, making it true.