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eimsori [14]
4 years ago
8

Pls someone help with this one

Mathematics
1 answer:
kupik [55]4 years ago
4 0

Answer:

490 feet

Step-by-step explanation:

You might be interested in
1. A cheetah can run at a top velocity of 31 m/s and has a mass of 47 kg. What is the cheetahs kenetic energy?
Naily [24]

Answer:

1. 22583.5 J

2. 1248 kg

3. 78.09 m/s

Step-by-step explanation:

The formula for kinetic energy is ;

K.E = 1/2* m * v²

Where m is the mass and v is the velocity

1.

K.E = 1/2 * m*v²

K.E = 1/2 * 47 * 31²

K.E = 22583.5 J

2.

Velocity = 9.78 m/s.

Kinetic energy = 60,800 J

To get mass, apply the formula as;

K.E = 1/2 * m*v²

60,800 = 1/2 * m * 9.78²

60800 = 48.70845 m

60800/48.70845 = m

1248 = m

Mass = 1248 kg

3.

K.E = 143.3 J

Mass = 0.047 kg

Apply the formula;

K.E = 1/2 * m*v²

143.3 = 1/2 * 0.047 * v²

143.3 = 0.0235 v²

143.3/0.0235 = v²

6097.87 = v²

v = √6097.87

v= 78.09 m/s

4 0
3 years ago
8.2.AP-5
miskamm [114]

Answer:

parallelogram

Step-by-step explanation:

Parallelograms have two pairs of parallel sides and no right angles.

8 0
3 years ago
The entires of the first row are
zubka84 [21]

Answer:

The answer is 3, 6.

Step-by-step explanation:

The numbers in the top rows on the left side of the equation is 1, 4 and 2, 2.

1, 4 + 1, 2 = (1 + 2), (4 + 2) = 3, 6

7 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
7x(15+5) plz help me solve this question by formula BODMAS
kupik [55]

The brackets are first, then the multiplication

7 0
3 years ago
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