Step-by-step explanation:
x = number hours dog walking
y = number hours tutoring
x + y <= 11
6x + 10y >= 80
the solution is the common area below the first line and above the second line.
the first line is above the second line (due to the y-intercept of 11 vs. 8) until the crossing point.
for the crossing point we use the regular equations :
from the first we get
x = 11 - y
using this in the second
6×(11 - y) + 10y = 80
66 - 6y + 10y = 80
4y = 14
y = 14/4 = 7/2 = 3.5
x = 11 - y = 11 - 3.5 = 7.5
so, she has to do at least 3.5 hours tutoring (and then 7.5 hours dog walking) for the minimum of $80 earning, all the way up to the maximum of 11 hours tutoring (and 0 dog walking) for $110 earning.
Answer:
22.9 feet
Step-by-step explanation:
You are using Tangent because you have the opposite and adjacent sides
Tan 5 = 2/x
x Tan 5= 2
x= 2/Tan5
= 22.860
(4 + 9 + 16 / 4) - 8 - (3 * 5)
(29 / 4) - 8 - (3 * 5) | Add up the numbers on left side.
(7.25) - 8 - (15) | Divide 29 by 4 and multiply 3 and 5.
-0.75 - (15) | Subtract 8 from 7.25
Final answer: -15.75
Answer:
<u><em>The Father is currently 47 and the Son is 7</em></u>
Step-by-step explanation:
Let F and S be the present ages of Father and Son, respectively.
We are told that <u>(F-2) = 9(S-2)</u> [2 years ago, father age was nine times the son age]
We also learn that <u>(F+3) = 5(S+3)</u> [3 years later it will be 5 times only]
Take the first expression and isolate one of the variables (S or F). I'll isolate F:
(F-2) = 9(S-2)
F = 9S - 16
Now use this in the second expression:
(F+3) = 5(S+3)
((9S-16)+3) = 5(S+3)
9S-13 = 5S+15
4S = 28
S = 7
Since F = 9S-16,
F = 9*(7)-16
F = 47
<u><em>Father is 47 and Son is 7</em></u>
CHECK:
Was the father 9 times the age of his son 2 years ago?
Father would have been 45 and son 5. Yes, 9*5 = 45
In 3 years will he be 5 times older than his son? Yes, Father would be 50 and son would be 10. 5*(10) = 50
