Using the speed - distance relationship, the time left before the appointed time is 27 minutes.
<u>Recall</u><u> </u><u>:</u>
<u>At</u><u> </u><u>10mph</u><u> </u><u>:</u>
- Distance = 10 × (t + 3) = 10t + 30 - - - (1)
<u>At</u><u> </u><u>12</u><u> </u><u>mph</u><u> </u><u>:</u>
- Distance = 12 × (t - 2) = 12t - 24 - - - - (2)
<em>Equate</em><em> </em><em>(</em><em>1</em><em>)</em><em> </em><em>and</em><em> </em><em>(</em><em>2</em><em>)</em><em> </em><em>:</em>
10t + 30 = 12t - 24
<em>Collect</em><em> </em><em>like</em><em> </em><em>terms</em><em> </em>
10t - 12t = - 24 - 30
-2t = - 54
<em>Divide</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>by</em><em> </em><em>-</em><em> </em><em>2</em>
t = 54 / 2
t = 27
Hence, the time left before the appointed time is 27 minutes.
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Let's solve your equation step-by-step.
(−
5
/8
)(x)=−160
Step 1: Simplify both sides of the equation.
−5
/8
x=−160
Step 2: Multiply both sides by 8/(-5).
(
8
/−5
)x(
−5
/8
x)=(
8/
−5
)*(−160)
x=256
Answer:
x=256
Area=pi times radius^2
area=36pi
36pi=pi times radius^2
divide both sides by pi
36=radius^2
square root both sides
6=radius
2radius=diameter
2 times 6=12
radius=6
diamater=12
the answers area A and D
(if you know your radius and diameter, you would know that diameter is 2 times radius and you could eliminate 2 choices imedietly (B and C))
Answer:
The equation
c x 2 > k
l x 3 ?> c > k
g = 30 = c + l + k
this equation is the same as...
Cecile read twice as many books as Kami.
Laura read 3 times as many books as Kami, but possibly less than Cecile.
Gigi read 30 books. 30 books is the same as all 3 girls combined.
To find out how many books each girl read, you need to find 3 unequal numbers that add up to 30. These numbers are substituted, only because they can be anything at all, and I want to prevent confusions.
hope this helps!
Surface area is the area that is on the outside of a figure and will be used when you have to cover something, but a volume is an area inside an object and is used in finding the amount someone can hold within something else.
