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levacccp [35]
3 years ago
13

renaldo catches the bus at 4:00 onto ride 3.2 miles from his house to the dentist office. He arrives at 4:30 p.m for one hour ap

pointment then he will ride a bus traveling at the same rate of speed 4.8 miles to the soccer feild. Will he be on time for his 6:30 p.m soccer practice? explain
Mathematics
1 answer:
SVEN [57.7K]3 years ago
7 0

Answer:

Yes, Renaldo will be reach on time for his Soccer practice.

Step-by-step explanation:

For traveling for 3.2 miles bus takes 30 minute times.

So, traveling a 1-mile bus will take 30 ÷ 3.2 minutes.

Now, Renaldo reaches the dentist's office at 4:30 p.m.

Then he has an appointment for 1 hour

So, Renaldo leaves the dentist's office at 5:30 p.m.

Now he has to travel 4.8 miles, it will take time = 4.8\times\frac{30}{3.2} = 45 minutes

So, Renaldo will reach Soccer field at

5 hours 30 minute + 45 minutes = 6 hours 15 minutes = 6:15 p.m

Hence, Renaldo will reach Soccer field before 6:30 p.m.

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For the function defined by y = 1/x4, y varies inversely as what quantity
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ANSWER

y varies inversely as x exponent 4.

EXPLANATION

The inverse variation equation is given as:

y =  \frac{1}{ {x}^{4} }

We can see that there is an inverse relation between the quantity y and x.

If the it were y =  \frac{1}{ {x}^2 }, we say y varies inversely as the square of x.

Hence for the given relation,the precise definition is that, y varies inversely as x exponent 4.

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If the numbers 4, 5 and 6 are each used exactly once to replace the letters in the expression A ( B − C ), what is the least pos
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Answer:

The least possible result is <em>-10</em>.

Step-by-step explanation:

Given the numbers 4, 5 and 6 are to be chosen one of the letters A, B or C.

First of all,

Let A = 4, B = 5 and C = 6

A(B-C) = 4 \times (5-6) = 4 \times -1 = -4

Let A = 4, B = 6 and C = 5

A(B-C) = 4 \times (6-5) = 4 \times 1 = 4

Let A = 5, B =4 and C = 6

A(B-C) = 5 \times (4-6) = 5 \times -2 = -10

Let A = 5, B = 6 and C = 4

A(B-C) = 4 \times (6-4) = 4 \times 2 = 8

Let A = 6, B = 4 and C = 5

A(B-C) = 6 \times (4-5) = 6 \times -1 = -6

Let A = 6, B = 5 and C = 4

A(B-C) = 6 \times (6-5) = 6 \times 1 = 6

Summarizing the above values in the form of a table:

\begin{center}\begin{tabular}{ c c c c}A & B & C & A(B-C)\\ 4 & 5 & 6 & -4\\  4 & 6 & 5 & 4\\  5 & 4 & 6 & -10\\  5 & 6 & 4 & 10\\  6 & 4 & 5 & -6\\ 6 & 5 & 4 & 6\end{tabular}\end{center}

So, the least possible result is <em>-10</em>.

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