Mary spent $10.30 on dinner to stay in budget.
Step-by-step explanation:
Amount spent on breakfast = $6.25
Amount spent on lunch = $8.95
Amount spent on dinner = x
Average = $8.50
Average = 

Multiplying both sides by 3

Mary spent $10.30 on dinner to stay in budget.
Keywords: Average, addition
Learn more about addition at:
#LearnwithBrainly
Answer:
Not correct
Step-by-step explanation:
Given:
x² - 6x - 7 = 0
The following equation can be solved correctly by factoring Thus,
x² - x + 7x - 7 = 0
x(x - 1) +7(x - 1) = 0
(x - 1)(x + 7) = 0
(x - 1) = 0
x - 1 = 0
x - 1 + 1 = 0 + 1
x = 1
Or
(x + 7) = 0
x + 7 = 0
x + 7 - 7 = 0 - 7
x = -7
Therefore, the solution would be x = 1 or x = -7
Lateena solution is therefore NOT CORRECT because she used a wrong method.
Answer:
A. 77.4
Step-by-step explanation:
<u>Linear approximation formula</u>

Given function:




Use the chain rule to differentiate the function.

Differentiate the two parts separately:


Put everything back into the chain rule formula:

.
The <u>linear approximation</u> at a = 8 is:

Finally, substitute x = 7.8 into the <u>linear approximation equation</u>:

Like 10% maybe your subtracting ight? message back so i can help