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3 years ago

9

Karl set out to Alaska on his truck. The amount of fuel remaining in the truck's tank (in liters) as a function of distance driv

en (in kilometers) is graphed.
What distance did Karl travel by the time he had 200 liters of fuel?

_____ kilometers

1 answer:

Stells [14]3 years ago

6 0

Answer:

500 kilometers

Step-by-step explanation:

The fuel (y) Karl has in his car is a function of the distance (x) Karl travelled. The amount of fuel (y) is dependent on the distance (x) Karl travelled.

On the y-axis, we have amount of fuel.

On the x-axis, we have distance travelled.

At the time Karl travelled when he had 200 liters of fuel (y) remaining in his car, he had travelled a distance (x) of 500 kilometers.

We know this by looking at the graph given. When x (distance) = 500, y (fuel) = 200.

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Solve the equations

ASHA 777 [7]

Answer:

(1)The value of x is -9 .

Option (B) is correct .

(2)The value of x is 13 .

Option (A) is correct .

(3)The value of the x is 11 .

Option (D) is correct .

(4) The value of the x is -15 .

Option (A) is correct .

(5)The value of the x is -11 .

Option (C) is correct .

Step-by-step explanation:

First Part

As given

4x = -36

$x = \frac{-36}{4}$

x = -9

Therefore the value of x is -9 .

Option (B) is correct .

Second Part

As given

5x - 15 = 50

5x = 50 + 15gg

5x = 65

$x = \frac{65}{5}$

x = 13

Therefore the value of x is 13 .

Option (A) is correct .

As given

4(x+2)-17=35

4x + 8 - 17 = 35

4x = 35 + 17 - 8

4x = 44

$x = \frac{44}{4}$

x = 11

Therefore the value of the x is 11 .

Option (D) is correct .

Fourth Part

As given

6x+12=4x-18

6x-4x = -12-18

2x = -30

$x = \frac{-30}{2}$

x = -15

Therefore the value of the x is -15 .

Option (A) is correct .

Fifth Part

As given

$\frac{1}{2}(4x-10)+5=x-11$

Simplify the above

4x-10+5×2 = 2x-22

4x-10+10= 2x-22

4x-2x = -22

2x=-22

$x = \frac{-22}{2}$

x = -11

Therefore the value of the x is -11 .

Option (C) is correct .

5 0

3 years ago

Read 2 more answers

Need help with compound fractions. Please answer both

Reika [66]

$\frac{8*( \frac{c}{c} )- \frac{3}{c} }{2*( \frac{7c}{7c})+ \frac{3}{7c} } \\ \\ \frac{ \frac{8c-3}{c} }{ \frac{14c+3}{7c} } \\ \\ \frac{8c-3}{c} * \frac{7c}{14c+3} \\ \\ = \frac{56c-21}{14c+3}$

Domain restrictions are what x values make the denominator zero.
a² + 5a - 36 = (a + 9)(a - 4)
a ≠ -9 ; a ≠ 4

6 0

2 years ago

In the circle below, DB = 22 cm, and m<DBC = 60°. Find BC. Ignore my handwriting.

Karo-lina-s [1.5K]

Answer:

$BC=11\ cm$

Step-by-step explanation:

step 1

Find the measure of the arc DC

we know that

The inscribed angle measures half of the arc comprising

$m\angle DBC=\frac{1}{2}[arc\ DC]$

substitute the values

$60\°=\frac{1}{2}[arc\ DC]$

$120\°=arc\ DC$

$arc\ DC=120\°$

step 2

Find the measure of arc BC

we know that

$arc\ DC+arc\ BC=180\°$ ----> because the diameter BD divide the circle into two equal parts

$120\°+arc\ BC=180\°$

$arc\ BC=180\°-120\°=60\°$

step 3

Find the measure of angle BDC

we know that

The inscribed angle measures half of the arc comprising

$m\angle BDC=\frac{1}{2}[arc\ BC]$

substitute the values

$m\angle BDC=\frac{1}{2}[60\°]$

$m\angle BDC=30\°$

therefore

The triangle DBC is a right triangle ---> 60°-30°-90°

step 4

Find the measure of BC

we know that

In the right triangle DBC

$sin(\angle BDC)=BC/BD$

$BC=(BD)sin(\angle BDC)$

substitute the values

$BC=(22)sin(30\°)=11\ cm$

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3 years ago

How many times does seven go into three hundred ninety four

Nookie1986 [14]

Answer: 56 times.

Step-by-step explanation:

394/7 = 56.3

56 * 7 = 392

392 + 7 = 399

399 is more than 394, therefore the answer is 56.

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2 years ago

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Mason, a hardwood flooring installer, needs to calculate the amount of hardwood flooring needed to cover a floor that's 33⁄4 yar

Aleksandr-060686 [28]

Answer:

20 5/8 yd^2

Step-by-step explanation:

width = 3 3/4 yd

length = 5 1/2 yd

First change both measurements to fractions.

3 3/4 = 3 + 3/4 = 12/4 + 3/4 = 15/4

5 1/2 = 5 + 1/2 = 10/2 + 1/2 = 11/2

area = length * width

area = 15/4 yd * 11/2 yd

area = (15 * 11)/(4 * 2) yd^2

area = 165/8 yd^2

165/8 = 20 remainder 5

area = 20 5/8 yd^2

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3 years ago