UGA2016
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**!!!STUCK ON 2 QUESTIONS!!!!

The time is takes to mow the lawn at a large park m(x) varies inversely with the numbers of workers assigned to the job x. It takes 90 minutes to complete the job when 3 workers are assigned to it.


which equation can be used to find the time to complete the job when x workers are assigned to it?

A.) m(x)= 270/x
B.) m(x) = 270x
C.) m(x) = 30/x
D.) m(x) = 30x

2. Suppose that H(x) varies inversely with x and H(x)=50 when x =0.25

What is H(x) when x =2?

A.) 0.5
B.) 6.25
C.) 12.5
D.) 24

Can you also provide how you got the answer as well :)

Respuesta :

The first one is A 
270/3=90 thus proving the equation right 

oh and the second one is B

Answer:

1. Which equation can be used to find the time to complete the job when x workers are assigned to it?

A.) m(x) = 270/x

2. Suppose that H(x) varies inversely with x and H(x)=50 when x =0.25

What is H(x) when x =2?

B.) 6.25

Step-by-step explanation:

1. We know when x=3 (workers), m(x)= 90 min (time is takes to mow the lawn at a large park) and we know m(x) varies inversely with x, this means when m(x) increases so x decreases. This is obtained with a division over x.

Now, we can replace to know what is the correct answer:

[tex]m(x)=\frac{y}{x} (1) \\90=\frac{y}{3} \\90*3=y[/tex]

[tex]y=270[/tex]

The answer is A.

[tex]m(x)=\frac{270}{x}[/tex]

We can confirm if we replace x=3 in the equation before. Our answer should be 90.

[tex]m(x)=\frac{270}{3}=90[/tex]

2.  Suppose that H(x) varies inversely with x and H(x)=50 when x =0.25

We know H(x) varies inversely with x, this means when H(x) increases so x decreases. This is obtained with a division over x. Now, we need to find the constant in the numerator.

Now, we can replace in the eq (1) to know the constant in the numerator:

[tex]H(x)=\frac{y}{x}\\50=\frac{y}{0.25} \\50*0.25=y[/tex]

[tex]y=12.5[/tex]

Now, we can replace in the eq (1) to find H(x) when x=2

[tex]m(x)=\frac{y}{x}[/tex]

[tex]m(x)=\frac{12.5}{2}[/tex]

[tex]m(x)=6.25[/tex]  

The answer is B.) 6.25

Q&A Education