Ask question

Ask question

All categories

3 years ago

9

A gas tank has a leak. Each hour it loses 3 gallons of gas. Write the equation that represents how much gas, g , the tank loses

over h hours.

1 answer:

zhenek [66]3 years ago

7 0

Let t is the amount of time that the gas is leaking (hour), g is the total of the gas leak.

g = 3t.

You might be interested in

Can anybody do this for<br> Me

Tanya [424]

Answer:

of which one we should provide ans

6 0

3 years ago

The map below shows the path that Jin walked this morning. According to his pedometer, the total distance of this path was 2.3 k

max2010maxim [7]

Answer:

5 centimetres = 1 kilometre

Step-by-step explanation:

First add up the distances measured on the map.

1.1+4.8+2.2+3.4=11.5 cm

Based on this, since the number of cm is greater than the number of km, we can figure out that each cm should not be multiplied to get the km, so we can eliminate the first two answers.

Next write an equation to represent finding the scale

x=scale

11.5=2.3x

Now solve for the scale

11.5/2.3=2.3x/2.3

5=x

Therefore the scale is 5 centimetres = 1 kilometre

6 0

3 years ago

The sum of the digits of a three-digit number is 16 .the tens digit is two more than the units digit and the unit digit is tripl

Anit [1.1K]

Answer:

286

Step-by-step explanation:

If the three-digit number is xyz, then:

x + y + z = 16

y = z + 2

z = 3x

Substitute the third equation into the second:

y = 3x + 2

Substitute this and the third equation into the first:

x + (3x + 2) + (3x) = 16

7x + 2 = 16

7x = 14

x = 2

y = 3x + 2 = 8

z = 3x = 6

The number is 286.

4 0

3 years ago

A cable hangs between two poles of equal height and 35 feet apart. At a point on the ground directly under the cable and x feet

gayaneshka [121]

Answer:

293.38 pounds

Step-by-step explanation:

We are given that

Distance between poles=35 feet

$h(x)=10+0.1(x^{1.5})$

Weight of cable=10.4 per linear foot

We have to find the weight of the cable.

Differentiate w.r.t

$h'(x)=0.1(1.5)x^{0.5}=0.15x^{0.5}$

$s=2\int_{0}^{17.5}\sqrt{1+(h'(x))^2}dx$

$s=2\int_{0}^{17.5}\sqrt{1+(0.15x^{0.5})^2}dx$

$s=2\int_{0}^{17.5}\sqrt{1+0.0225x}dx$

Let $1+0.0225x=t$

$dx=\frac{1}{0.0225}dt$

$s=\frac{2}{0.0225}\int_{0}^{17.5}\sqrt{t}dt$

$s=\frac{2}{0.0225}\times\frac{2}{3}[t^{\frac{3}{2}}]^{17.5}_{0}$

$s=2\times \frac{2}{3\times0.0225}[(1+0.0255x)^{\frac{3}{2}]^{17.5}_{0}$

$s=\frac{4}{3\times 0.0225}((1+0.0225(17.5))^{\frac{3}{2}-1)$

s=28.21

Weight of cable= $28.21\times 10.4=293.38$ pound

8 0

3 years ago

Put these questions in order from least to greatest 3/8 19/24 2/3

aleksandr82 [10.1K]

First 3/8, second 2/3, and third 19/24

4 0

3 years ago