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2 years ago

Could some please help me with this?

Mathematics

2 answers:

disa [49]2 years ago

8 0

C: 20 m/s

Explanation:
Distance run: 1000m
Time passed: 50s
Distance divided by time equals speed.

1000m / 50s = 20 m/s

Vladimir79 [104]2 years ago

4 0

Answer: C

Step-by-step explanation:

S(speed)=D(distance)/T(time) so (d)1000/(t)50= 20m/s

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BRAINLIEST + POINTS! Help needed and explanation please!

sattari [20]

The Median is the middle value.

Arrange the numbers in order from smallest to largest:

Seattle: 23.78 , 32.46, 40.30, 45.79, 49.42, 55.14

Since there is an even number, the median is found by adding the 2 middle numbers and dividing by 2.

Median for Seattle: 40.30 + 45.79 = 86.09 / 2 = 43.045

Now do the same for Las Vegas:

2.33, 3.72, 3.84, 4.11, 5.62, 8.79

Median for Las Vegas: 3.84 + 4.11 = 7.95 / 2 = 3.975

Difference: 43.045 - 3.975 = 39.07

The answer is B.

5 0

3 years ago

Read 2 more answers

Which of the values is equivalent to 5^–3? A) 1/15 B) 1/125 C) − 15 D) − 125

bazaltina [42]

Answer:

C 1/125

Step-by-step explanation:

5 ^−3

Calculate 5 to the power of −3 and get

1/125

7 0

2 years ago

Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature is 45 degrees at midnig

Nady [450]

Answer:

$D = 45+19*sin(\frac{\pi}{12}*t)$

Step-by-step explanation:

The amplitude (A) and the mean (M) temperature are given by:

$A=\frac{64-26}{2}\\ A= 19\\M=\frac{64+26}{2}\\ M= 45\\$

The mean temperature, 45 degrees. occurs at midnight (t=0) and the frequency is f=24h. Therefore, the temperature D, in degrees, as a function of t, in hours, is:

$D = M+A*sin(\frac{2\pi}{24}*t)\\D = 45+19*sin(\frac{\pi}{12}*t)$

5 0

3 years ago

For each pair figures, find the ratio of the area of the first figure to the area of the second. 14mm 7mm

Paraphin [41]

The ratio of the area of the first figure to the area of the second figure is 4:1

<h3>Ratio of the areas of similar figures </h3>

From the question, we are to determine the ratio of the area of the first figure to the area of the second figure

The two figures are similar

From one of the theorems for similar polygons, we have that

If the scale factor of the sides of two similar polygons is m/n then the ratio of the areas is (m/n)²

Let the base length of the first figure be ,m = 14 mm

and the base length of the second figure be, n = 7 mm

∴ The ratio of their areas will be

$(\frac{14 \ mm}{7 \ mm})^{2}$

$= \frac{196 \ mm^{2} }{49\ mm^{2} }$

$=\frac{4}{1}$

= 4:1

Hence, the ratio of the area of the first figure to the area of the second figure is 4:1

Learn more on Ratio of the areas of similar figures here: brainly.com/question/11920446

8 0

1 year ago

Below the function h (x) ang g (x) are graphes. At which values of x is true that h (x) > g (x)? Select all that apply.

rodikova [14]

Answer:

Step-by-step explanation:

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2 years ago