This question s incomplete, the complete question is;
The Watson family and the Thompson family each used their sprinklers last summer. The Watson family's sprinkler was used for 15 hours. The Thompson family's sprinkler was used for 30 hours.
There was a combined total output of 1050 of water. What was the water output rate for each sprinkler if the sum of the two rates was 55L per hour
Answer:
The Watson family sprinkler is 40 L/hr while Thompson family sprinkler is 15 L/hr
Step-by-step explanation:
Given the data in the question;
let water p rate for Watson family and the Thompson family sprinklers be represented by x and y respectively
so
x + y = 55 ----------------equ1
x = 55 - y ------------------qu2
also
15x + 30y = 1050
x + 2y = 70 --------------equ3
input equ2 into equ3
(55 - y) + 2y = 70
- y + 2y = 70 - 55
y = 15
input value of y into equ1
x + 15 = 55
x = 55 - 15
x = 40
Therefore, The Watson family sprinkler is 40 L/hr while Thompson family sprinkler is 15 L/hr
I think it's 8 b/c 24 minutes divided by 3 miles is 8 minutes per mile.
Answer:
D
Step-by-step explanation:
Answer:
Height of flagpole = 571 ft
Step-by-step explanation:
Given:
Angle of elevation to the top of the flagpole is 55° .
Distance from the eyes to the base of flagpole = 400 ft.
To find the height of the flagpole.
Solution:
We can draw the situation as a right triangle as shown below.
In triangle ABC.
∠C= 55°
To find the length AB (height of the flagpole).
Applying trigonometric ratio :
Plugging in values.
Multiplying both sides by 400.
∴ 
Height of flagpole = 571 ft
6.5X + 9(30) = 8(30+X)6.5X +270 = 240 + 8X270-240 = 8X-6.5X1.5X = 30X = 30/1.5 =20 ounces of $6.50 alloy9 is within 1 of 86.5 is within 1.5 of 81.5 is 3/2 of 130 is 3/2 of 20You know you need more of $9 alloy since it's closer to $8You need exactly 3/2 or 1.5 times more of the $9 alloy30 is 1.5 times 20
