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Masja [62]
2 years ago
6

Rob runs at a constant speed of 3.7 meters per second for half an hour. Then, he walks at a rate of 1.5 meters per second for ho

ur. How far did Rob run and walk in 60 minutes?
Rob ran and walked ? meters. ​
Mathematics
1 answer:
Eva8 [605]2 years ago
7 0

Answer:

Let us look at the problem, shall we?

Step-by-step explanation:

We know that rob runs 3.7 meters per second so 3.7 meters per second x 60 for 1 minute = 222 meters times 30 for 30 minutes= 6660 meters. Now 1.5 meters per second,= 90 meters per second x 60=5400 + 6660= 12060

(I might be wrong)

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2. Which expression is equivalent to 5(n + 3) - 10? A. 5+1 +5+3 – 10 B. 5 • n + 5 + 3 - 10 C. 5. n + 5 • 3 – 10 D. 5 • n • 5 • 3
GuDViN [60]

Answer:

5n+5

Step-by-step explanation:

5(n+3)-10

5n+15-10

5n+5

8 0
3 years ago
Remove the brackets using the distributive law for the following: 2(x + 3)
Vanyuwa [196]

Answer:

2x + 6

Step-by-step explanation:

= 2(x + 3)

= 2(x) + 2(3)

= 2x + 6

5 0
3 years ago
In the spinner below the large wedges are twice the size of the smaller ones. What is true about the probability of landing on 6
mr_godi [17]

Answer:

A

Step-by-step explanation:

If we take each small section to be "1" unit, we can say the large sections (for "2" and "5") are "2" units each. So in total there will be 8 sections.

Since 5 is "2" sections, we can say:

P(5) = 2/8 = 1/4

And 6 is "1" section, so we can say:

P(6) = 1/8

Definitely, Probability of landing a 6 is HALF that of probability of landing a 5. Also we can see this is the picture.

So, from the answer choices, A is right.

6 0
3 years ago
Write 17/9 as a decimal
denis23 [38]
Divide 17 by 9
1.8888
6 0
3 years ago
Read 2 more answers
Write the integral that gives the length of the curve y = f (x) = ∫0 to 4.5x sin t dt on the interval ​[0,π​].
Troyanec [42]

Answer:

Arc length =\int_0^{\pi} \sqrt{1+[(4.5sin(4.5x))]^2}\ dx

Arc length =9.75053

Step-by-step explanation:

The arc length of the curve is given by \int_a^b \sqrt{1+[f'(x)]^2}\ dx

Here, f(x)=\int_0^{4.5x}sin(t) \ dt interval [0, \pi]

Now, f'(x)=\frac{\mathrm{d} }{\mathrm{d} x} \int_0^{4.5x}sin(t) \ dt

f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( [-cos(t)]_0^{4.5x} \right )

f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( -cos(4.5x)+1 \right )

f'(x)=4.5sin(4.5x)

Now, the arc length is \int_0^{\pi} \sqrt{1+[f'(x)]^2}\ dx

\int_0^{\pi} \sqrt{1+[(4.5sin(4.5x))]^2}\ dx

After solving, Arc length =9.75053

5 0
3 years ago
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