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3 years ago

Find the inverse variation equation that relates x and y.

Mathematics

1 answer:

lakkis [162]3 years ago

8 0

DONT CLICK THE LINK ITS A SCAM!!! Just letting you know!

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What represent the relationship between the total mass of a crate

sineoko [7]

9514 1404 393

Answer:

(a) M = 0.25n +100

Step-by-step explanation:

The distance between the dots on the graph is a rise of 1 grid square and a run of 2 grid squares. If we extend the sequence of dots to the left, we expect to place one at (0, 100). That is, the y-intercept of this function is 100 (eliminates choices C and D).

The rise of 1 grid square represents 25 kg, and the run of 2 grid squares represents 100 CDs. Then the slope of the function (rate of change) is ...

slope = rise/run = 25/100 kg/CD = 0.25

Then the equation describing the points on the graph will be ...

M = 0.25n +100

5 0

3 years ago

A 4 foot man cast a 6 foot shadow how tall is a tree that cast a 21 foot shadow

Eva8 [605]

Answer:

Step-by-step explanation:

Maybe 8 feet

3 0

2 years ago

Read 2 more answers

Nina has 3 feet ribbon that she wants to cut into 8 equal sized pieces. What is true about the length of each ribbon? A) each pi

tresset_1 [31]

B)each piece has 1/8 of 3 feet ribbon

8 0

3 years ago

Read 2 more answers

If G(x) = 3x + 1, then G -1(1) is<br> 4<br> 1/3<br> 0

Eduardwww [97]

Answer:

Step-by-step explanation:

$g(x) = 3x + 1 \\\ y = 3x + 1$

Changing x to y and y to x:

$x = 3y + 1 \\\ 3y = x - 1 \\\ y = \frac{x - 1}{3} \\\ g^{-1}(x) = \frac{x - 1}{3} \\\ g^{- 1} (1) = \frac{1 - 1}{3} \\\ g^{- 1} (1) = 0$

I hope I helped you.

4 0

3 years ago

równanie zmiany prędkość autokaru poruszającego się po prostym odcinku szosy i rozpoczynającego hamowanie od szybkości 20m/s ma

Rasek [7]

Answer:

s = 22.5 m

Step-by-step explanation:

the equation for the speed change of a coach moving along a straight section of the road and starting braking at a speed of 20 m / s has the form v (t) = 25-5t. Using integral calculus, determine the coach's braking distance.

v (t) = 25 - 5 t

at t = 0 , v = 20 m/s

Let the distance is s.

$s =\int v(t) dt\\\\s =\int (25 - 5t)dt\\\\s= 25 t - 2.5 t^2 \\$

Let at t = t, the v = 20

So,

20 = 25 - 5 t

t = 1 s

So, s = 25 x 1 - 2.5 x 1 = 22.5 m

3 0

2 years ago