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3 years ago

11

How do you know how long a trip will take if you have 399 mileage on one tank of gas but the maximum you can drive in one day is

600?

1 answer:

andrey2020 [161]3 years ago

3 0

Answer:

Step-by-step explanation:

600 minus 399 is 201

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Please help me with this

inn [45]

It's the first one and the last one need any more help let me know okay

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2 years ago

Ericka decided to compare her observation to the average annual trend, which shows the water rising 1.8 mm/year. Remember, she u

Alex787 [66]

Answer with explanation:

Presume that, initial water level to be, 0 mm, at a time t=0.

And water level is rising at the rate of $1.8 \frac{\text{mm}}{\text{year}}$ .

After 6.2 years level of water will be =1.8 × 6.2

=11.16 mm

→Averge, that is increase in water level after 6.2 years is given by

$=\frac{W_{2}-W_{1}}{t_{2}-t_{1}}\\\\=\frac{11.16-0}{6.2-0}\\\\=\frac{11.16}{6.2}\\\\=1.8 \frac{\text{mm}}{\text{year}}$

→→Rate of Depreciating of water level is given by = Negative of Increase of water level= $-1.8 \frac{\text{mm}}{\text{year}}$

5 0

3 years ago

Read 2 more answers

Please help me with this<br><br> a . 18<br><br> b . 16<br><br> c . 12<br><br> d . 9

lbvjy [14]

It would have the be B. 16

8 0

2 years ago

Let V be the volume of the solid obtained by rotating about the y-axis the region bounded y = 4x and y = x2/4 . Find V by slicin

aliya0001 [1]

Answer:

Step-by-step explanation:

Consider the graphs of the $y = 4x$ and $y = \frac{x^{2} }{4}$ .

By equating the expressions, the intersection points of the graphs can be found and in this way delimit the area that will rotate around the Y axis.

$4x = \frac{x^{2} }{4} \\ x^{2} = 16x \\ x^{2} - 16x = 0 \\ x(x-16) = 0$ then $x=0$ o $x=16$ . Therefore the integration limits are:

$y = 4(0) = 0$ and $y = 4(16) = 64$

The inverse functions are given by:

$x = 2 \sqrt{y}$ and $x = \frac{y}{4}$ . Then

The volume of the solid of revolution is given by:

$\int\limits^{64}_ {0} \, [2\sqrt{y} - \frac{y}{4}]^{2} dy = \int\limits^{64}_ {0} \, [4y - y^{3/2} + \frac{y^{2}}{16} ]\ dy = [2y^{2} - \frac{2}{5}y^{5/2} + \frac{y^{3}}{48} ]\limits^{64}_ {0} = 546.133 u^{2}$

6 0

3 years ago

A rectangular flower bed that is 5 ft long and 4 feet wide is surrounded by a square lawn that is

mr Goodwill [35]

Answer:

124 ft2

Step-by-step explanation:

The yard is a 12ft square so the length and width would both be 12ft. You multiply the length and width of the yard to get the area of the yard so 12x12=144. You have the length and width of the flower bed so 5x4=20. Once you have the area of both you can subtract the area the flower bed takes up of the lawn. 144-20=124 ft2.

6 0

3 years ago