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3 years ago

5

The gallons of gas varies directly to number hours driving a car. Dylan drove 3 hours on 15 gallons of gas. If he drove for 35 h

ours, how many gallons of gas is that?

2 answers:

Arte-miy333 [17]3 years ago

8 0

Answer:

I am unable to solve this problem Sorry

Step-by-step explanation:

In-s [12.5K]3 years ago

5 0

Answer:

About 30

Step-by-step explanation:

I Multiply 3 to 35 then subtracted 15

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Graph the numbers that are solutions to both inequalities below.

aleksklad [387]

Answer:

3x>12 2x/3 ≤ 6

3x>12, x = 4 2x/3, x = 2 ≤ 6

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2 years ago

Robin earned $4,320 in 9 weeks and worked 40 hours in each week. How much did he earn per hour?

quester [9]

Divide 4,320 between 9 and you'll get 408 per week. divide 408 between 40 and you'll get how much he makes per hour

6 0

3 years ago

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A spherical balloon is inflated so that its volume is increasing at the rate of 3 ft3/min. How fast is the diameter of the ballo

beks73 [17]

When an spherical balloon volume is increasing at the rate of $3ft^3/min$ then the diameter of the balloon is increasing $\frac{3}{2\pi }ft /min$

How can we find the rate of change of balloon's diameter ?

The volume of a spherical balloon is $v=\frac{4}{3} \pi r^3$

In form of diameter we can write as

$v=\frac{4}{3} \pi (\frac{D}{2} )^3\\=\frac{1}{6} \pi D^3$

Now we will differentiate both sides wrt to $t$ we get

$\frac{dv}{dt} =\frac{1}{6} \pi 3D^2 \frac{dD}{dt} \\\frac{dD}{dt} =\frac{2}{\pi D^2} \frac{dv}{dt} \\\\when r=1\\D=2ft$

Given in the question $\frac{dv}{dt} =3ft^3/min$

thus when we substitute the values we get

$\frac{dD}{dt} =\frac{2}{\pi *2^2} (3)\\\frac{dD}{dt}=\frac{3}{2\pi } ft/min$

Learn more about the differentiation here:

brainly.com/question/28046488

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2 years ago

Simplify (9x + 10) - (7x + 8)

Ilia_Sergeevich [38]

Answer:

2x+2

Step-by-step explanation:

of symbols, representing a value, o relation. example 2+2=4

8 0

3 years ago

Please let me know ASAP

aksik [14]

Answer:

∠ BAC = 80°

Step-by-step explanation:

The sum of the interior angles of quadrilateral ACDB = 360°

DB and DC are tangents to the circle, thus

∠ DBA = ∠ ACD = 90° ( angle between tangent/ circle at point of contact )

Thus

∠ BAC + 90° + 90° + 100° = 360°

∠ BAC + 280° = 360° ( subtract 280° from both sides )

∠ BAC = 80°

4 0

3 years ago

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