62 inchezzzzzzz so 60-4 = 56 --- 56 +6 = 62
The answer to the question is B.
Answer:
- one orange costs $0.35 and one grapefruit costs $0.45
Step-by-step explanation:
<em>let grapefruits be g , let oranges be o</em>
so first equation:
→2g + 3o = $1.95
→g = ($1.95 - 3o)/2
so second equation:
→ 3g + 2o = $2.05
→ g = ($2.05 - 2o)/3
Solving simultaneously:
→ ($2.05 - 2o)/3 = ($1.95 - 3o)/2
→ 2($2.05 - 2o) = 3($1.95 - 3o)
→ 4.1 - 4o = 5.85 - 9o
→ - 4o + 9o = 5.85 - 4.1
→ 5o = 1.75
→ o = $0.35
Therefore one orange costs $0.35
Then one grapefruit costs g → ($1.95 - 3o)/2
→ g = ($1.95 - 3(0.35))/2
→ g = 0.45
Therefore one grapefruit costs $0.45
Answer:
C. 4 times x squared plus 6 times x minus 3 minus 12 over the quantity 2 times x plus 1
Step-by-step explanation:
f(x) = 8x^3 + 16x^2 − 15
g(x) = 2x + 1.
f(x) / g(x) = (8x^3 + 16x^2 − 15) / (2x + 1)
Using long division
Answer:
60 minutes for the larger hose to fill the swimming pool by itself
Step-by-step explanation:
It is given that,
Working together, it takes two different sized hoses 20 minutes to fill a small swimming pool.
takes 30 minutes for the larger hose to fill the swimming pool by itself
Let x be the efficiency to fill the swimming pool by larger hose
and y be the efficiency to fill the swimming pool by larger hose
<u>To find LCM of 20 and 30</u>
LCM (20, 30) = 60
<u>To find the efficiency </u>
Let x be the efficiency to fill the swimming pool by larger hose
and y be the efficiency to fill the swimming pool by larger hose
x = 60/30 =2
x + y = 60 /20 = 3
Therefore efficiency of y = (x + y) - x =3 - 2 = 1
so, time taken to fill the swimming pool by small hose = 60/1 = 60 minutes
