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aleksandr82 [10.1K]
3 years ago
11

Work the following word problem. Set up the problem in numerical form before you solve it. Then find and reduce the answer. How

many times will 1/5 go into 37
Mathematics
2 answers:
FromTheMoon [43]3 years ago
6 0

Answer:

185 times of \dfrac{1}{5} go into 37.

Step-by-step explanation:

Consider the provided information.

How many times will 1/5 go into 37

First setup the problem in numerical form.

Let x denote the required number of times.

Then, according to the given information, we have

x\times \dfrac{1}{5}=37\\\\\\\Rightarrow x=5\times37\\\\\Rightarrow x=185.

Thus, 185 times of \dfrac{1}{5} go into 37.

OLga [1]3 years ago
3 0

Answer:

I believe the answer would be 7.4 times.

Step-by-step explanation:

37*.20=7.4

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A ball is thrown up in the air, and it's height (in feet) as a function of time (in seconds) can be written as h(t) = -16t2 + 32
olga2289 [7]

Answer:

1) Using properties of the quadratic equation:

Here the height equation is:

h(t) = -16t^2 + 32t + 6

We can see that the leading coefficient is negative, this means that the arms of the graph will open downwards.

Then, the vertex of the quadratic equation will be the maximum.

Remember that for a general quadratic equation:

a*x^2 + b*x + c = y

the x-value of the vertex is:

x = -b/2a

Then in our case, the vertex is at:

t = -32/(2*-16) = 1

The maximum height will be the height equation evaluated in this time:

h(1) = -16*1^2 + 32*1 + 6 = 22

The maximum height is 22ft

2) Second method, using physics.

We know that an object reaches its maximum height when the velocity is equal to zero (the velocity equal to zero means that, at this point, the object stops going upwards).

If the height equation is:

h(t) = -16*t^2 + 32*t + 6

the velocity equation is the first derivation of h(t)

Remember that for a function f(x) = a*x^n

we have that:

df(x)/dx = n*a*x^(n-1)

Then:

v(t) = dh(t)/dt = 2*(-16)*t + 32 + 0

v(t) = -32*t + 32

Now we need to find the value of t such that the velocity is equal to zero:

v(t) = 0 = -32*t + 32

       32*t = 32

            t = 32/32 = 1

So the maximum height is at t = 1

(same as before)

Now we just need to evaluate the height equation in t = 1:

h(1) = -16*1^2 + 32*1 + 6 = 22

The maximum height is 22ft

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3 years ago
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Answer: Lol ten points for your whole assignment

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