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2 years ago

13

Freddy went to the museum and bought a reduced copy of a painting he saw in a exhibit. If the actual painting is 12 inches high,

how wide is it ?

1 answer:

Snowcat [4.5K]2 years ago

4 0

Answer:

It is probably 6 inches wide. I tried my best

Step-by-step explanation:

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Who is able to pass out flyers at a faster rate

Alexxandr [17]

Answer:

I don't know because you did not provide enough info

Step-by-step explanation:

but personally I think that I could pass the flyers out the fastest

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3 years ago

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Make a conjecture about the next item in the sequence<br> 1, 1, 2, 3, 5, 8

Marizza181 [45]

It would be 13 because you are adding 5 and 8

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3 years ago

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PLEASE HELP ME WITH THIS QUESTION what is -48-(-12)

olga_2 [115]

Answer:

-36

Step-by-step explanation:

-48+12=-36

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3 years ago

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The distance d in miles that can be covered by a rabbit in X hours is given in the equation y = 30 x The distance covered by a b

AveGali [126]

Answer:

Rabbit is faster than bear.

Step-by-step explanation:

Speed of rabbit or bear is determined by their rate of change of their distances with time.

The distance covered by rabbit in $x$ hours is given as:

$y=30x$

The above equation is of the form $y=mx$ , where, $m$ is the rate of change of $y$ with $x$ .

Therefore, the rate of change of rabbit's distance with time is 30 mi/h.

Now, from the graph, the slope is given as the rise over run.

Rise is 50 miles and run is 2 miles.

So, rate of change of bear's distance with time is, $m_b=\frac{50}{2}=25\ mi/h$

Now, 30 is greater than 25. So, rabbit is faster than bear.

6 0

3 years ago

In basketball, hang time is the time that both of your feet are off the ground during a jump. The equation for hang time is t =

const2013 [10]

Considering the hang time equation, it is found that Player 1 jumped 0.68 feet higher than Player 2.

<h3>What is the hang time equation?</h3>

The hang-time of the ball for a player of jump h is given by:

$t = 2\left(\frac{2h}{32}\right)^{\frac{1}{2}}$

The expression can be simplified as:

$t = 2\sqrt{\frac{h}{16}}$

For a player that has a hang time of 0.9s, the jump is found as follows:

$0.9 = 2\sqrt{\frac{h}{16}}$

$\sqrt{\frac{h}{16}} = \frac{0.9}{2}$

$(\sqrt{\frac{h}{16}})^2 = \left(\frac{0.9}{2}\right)^2$

$h = 16\left(\frac{0.9}{2}\right)^2$

h = 3.24 feet.

For a player that has a hang time of 0.8s, the jump is found as follows:

$0.8 = 2\sqrt{\frac{h}{16}}$

$\sqrt{\frac{h}{16}} = \frac{0.8}{2}$

$(\sqrt{\frac{h}{16}})^2 = \left(\frac{0.8}{2}\right)^2$

$h = 16\left(\frac{0.8}{2}\right)^2$

h = 2.56 feet.

The difference is given by:

3.24 - 2.56 = 0.68 feet.

More can be learned about equations at brainly.com/question/25537936

#SPJ1

3 0

2 years ago