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4 years ago

15

Yeah so 5mi/h =?km/h

1 answer:

tamaranim1 [39]4 years ago

8 0

Answer:

80 kilometers

Step-by-step explanation:

Since the number of miles and number of km is over one hour, the only thing you need to worry about it the number of miles in a kilometer (the numerator).

1 mile is approximately 1.60934 kilometers, so to find how many kilometers are in 50 miles, you would multiply 1.60934 by 50.

1.60934 * 50 = 80.467

Since 50 miles is rounded to the nearest whole number, I expect that the kilometers should be given in the nearest whole number, so the final answer is 80 kilometers.

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Is 2.4444 rational or irrational

antiseptic1488 [7]

Answer:

it has to be irrational

4 0

3 years ago

Read 2 more answers

Which number is located between 15 and 20 on a number line?

jeyben [28]

Answer: 17.5 is located between them.

Step-by-step explanation:

To find out which number is located between the numbers, find their mean average which sum of the numbers divide by the number of values.

15 + 20 = 35

35 / 2 = 17.5

8 0

3 years ago

Determine the common ratio and find the next three terms of the geometric sequence 9,3sqrt3,3

Maru [420]

Answer:

Fourth term: $a_4 = 9 * (\frac{\sqrt{3}}{3})^{(4 - 1)}$ = $9 * (\frac{\sqrt{3}}{3})^{3}$ = $\sqrt{3}$

Fifth term: $a_5 = 9 * (\frac{\sqrt{3}}{3})^{(5 - 1)}$ = $9 * (\frac{\sqrt{3}}{3})^{4}$ = 1

Sixth term: $a_6 = 9 * (\frac{\sqrt{3}}{3})^{(6 - 1)} = 9 * (\frac{\sqrt{3}}{3})^{5} =\frac{\sqrt{3}}{3}$

Step-by-step explanation:

The geometric progression is:

$9, 3 \sqrt{3}, 3...$

The first term, a, is 9

To find the common ratio, r, all we have to do is divide a term by its preceding term.

Let us divide the second term by the first:

$r = \frac{3\sqrt{3}}{9}\\ \\r = \frac{\sqrt{3}}{3}$

That is the common ratio.

Geometric progression is given generally as:

$a_n = ar^{(n - 1)}$

where a = first term

r = common ratio

$a_n$ = nth term

We need to find the 4th, 5th and 6th terms.

Fourth term: $a_4 = 9 * (\frac{\sqrt{3}}{3})^{(4 - 1)}$ = $9 * (\frac{\sqrt{3}}{3})^{3}$ = $\sqrt{3}$

Fifth term: $a_5 = 9 * (\frac{\sqrt{3}}{3})^{(5 - 1)}$ = $9 * (\frac{\sqrt{3}}{3})^{4}$ = 1

Sixth term: $a_6 = 9 * (\frac{\sqrt{3}}{3})^{(6 - 1)} = 9 * (\frac{\sqrt{3}}{3})^{5} =\frac{\sqrt{3}}{3}$

5 0

3 years ago

A family tree business removes overgrown trees for 3 months in the summer. In June, they removed 1/3 the number of trees as they

devlian [24]

Answer: 299

Step-by-step explanation:

Given

Family removes one-third of tress in June as much as they dd in July

They remove twice the number of July in august

Suppose they removed x trees in July

for June it is, $\frac{x}{3}$

for August, it is $2x$

If they removed less than 1000 tress in 3 months

$\Rightarrow \dfrac{x}{3}+x+2x$

So, maximum number of tress removed in July is 299

7 0

3 years ago

How many solutions exist for the given equation?

anygoal [31]

Answer:

None

Step-by-step explanation:

If you graph the points on a plot, they would be parallel

6 0

3 years ago