A line has a slope of 1/7 and includes the points (0,–4) and (u,–5). What is the value of u?
2 answers:
Answer is u= -7
Step by step
We know point (0, -4) which means our y intercept is at -4
We know slope is 1/7
y = 1/7x - 4
Graph it and find where the line crosses the x axis at -5 y which is -7 so our point is (-7, -5)
You can also use the formula
(y2 - y1) over (x2 - x1)
-5 - -4 over “u” - 0
-1 over ( u - 0 = -7 so u = -7)
-1/-7 = 1/7 slope
Step by step
We know point (0, -4) which means our y intercept is at -4
We know slope is 1/7
y = 1/7x - 4
Graph it and find where the line crosses the x axis at -5 y which is -7 so our point is (-7, -5)
You can also use the formula
(y2 - y1) over (x2 - x1)
-5 - -4 over “u” - 0
-1 over ( u - 0 = -7 so u = -7)
-1/-7 = 1/7 slope
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Hello,
Please, see the attached files.
Thanks.
Please, see the attached files.
Thanks.
Answer:
w= 9
Step-by-step explanation:
Square both sides:
-4w +61= (w -4)²
Expand:
-4w +61= w² -2(w)(4) +4²
-4w +61= w² -8w +16
Simplify:
w² -8w +16 +4w -61= 0
w² -4w -45= 0
Factorize:
(w -9)(w +5)= 0
w -9= 0 or w +5= 0
w= 9 or w= -5 (reject)
Note:
-5 is rejected since we are only taking the positive value of the square root here. If the negative square root is taken into consideration, then w= -5 would give us -9 on both sides of the equation.
<u>Why </u><u>do </u><u>we </u><u>use </u><u>negative </u><u>square </u><u>root?</u>
When solving an equation such as x²= 4,
we have to consider than squaring any number removes the negative sign i.e., the result of a squared number is always positive.
In the case of x²= 4, x can be 2 or -2. Thus, whenever we introduce a square root, a '±' must be used.
However, back to our question, we did not introduce the square root so we should not consider the negative square root value.