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3 years ago

A bus can carry 50 people per run. At least how many times does the bus have to run in order to transfer 25, 000 people?

Mathematics

1 answer:

galina1969 [7]3 years ago

7 0

50 × 500 = 25,000

The answer is 500 runs.

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Solve −2(−4n + 1)+5(25n − 8)=−62. <br><br><br><br> n= ?

In-s [12.5K]

N= -20/133

Step 1: Simplify both sides of the equation.
−
2
(
−
4
n
+
1
)
+
5
(
25
n
−
8
)
=
−
62
(
−
2
)
(
−
4
n
)
+
(
−
2
)
(
1
)
+
(
5
)
(
25
n
)
+
(
5
)
(
−
8
)
=
−
62
(Distribute)
8
n
+
−
2
+
125
n
+
−
40
=
−
62
(
8
n
+
125
n
)
+
(
−
2
+
−
40
)
=
−
62
(Combine Like Terms)
133
n
+
−
42
=
−
62
133
n
−
42
=
−
62
Step 2: Add 42 to both sides.
133
n
−
42
+
42
=
−
62
+
42
133
n
=
−
20
Step 3: Divide both sides by 133.
133
n
133
=
−
20
133
n
=
−
20
133

5 0

2 years ago

Read 2 more answers

I need help it needs to be done today but i dont know the answer

GrogVix [38]

Do you have an image?

4 0

3 years ago

Read 2 more answers

an ellipse has a center at the origin , a vertex along the major axis at (13,0), and a focus at (12,0). What is the equation of

-BARSIC- [3]

Answer:

The answer to your question is below

Step-by-step explanation:

Data

Center = (0, 0)

Vertex = (13, 0)

Focus = (12, 0)

Process

From the data we know that it is a horizontal ellipse.

1.- Calculate "a", the distance from the center to the vertex.

a = 13

2.- Calculate "c", the distance from the center to the focus

c = 12

3.- Calculate b

Use the Pythagorean theorem to find it

a² = b² + c²

-Solve for b

b² = a² - c²

-Substitution

b² = 13² - 12²

-Simplification

b² = 169 - 144

b² = 25

b = 5

4.- Find the equation of the ellipse

$\frac{x^{2} }{13^{2}} + \frac{y^{2}}{5^{2}} = 1$ or $\frac{x^{2} }{169} + \frac{y^{2}}{25} = 1$

7 0

3 years ago

Can you help me please

brilliants [131]

Answer:

WOW

Step-by-step explanation:

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3 0

3 years ago

a young hockey player on frictionless ice shoots a 1 kg hockey puck 150 m/s toward the goal. of the hockey player slides backwar

Dafna11 [192]

By conservation of linear amount of movement we have:
m1v1i + m2v2i = m1v1f + m2v2f
Where,
v1i = v2i = 0 (assuming initial rest)
m1v1f + m2v2f = 0
Substituting the values:
m1 (-3) + (1) (150) = 0
Clearing m1:
m1 = (- (1) * (150)) / (- 3)
m1 = 50 Kg
answer:
his mass is 50 Kg

5 0

3 years ago