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olga nikolaevna [1]
4 years ago
7

The distance between the Sun and the Earth is 1 .496 x 1O^8km and distance between the Earth and lhe Moon is 3.84 x 10^8m. Durin

g solar eclipse. The Moon comes in between the Earth and the Sun. What is the distance between the Moon and the Sun at that particular time?
Mathematics
1 answer:
Dennis_Churaev [7]4 years ago
3 0

Answer:

1.49216*10^8 km

Step-by-step explanation:

The first thing is to pass all the distances to the same unit system, since the distance from the earth to the sun is in kilometers and from the earth to the moon is in meters, therefore we will pass the distance of the moon in kilometers:

3.84 x 10 ^ 8m * 1 km / 1000 m = 3.84 x 10 ^ 5 km

To calculate the distance between the moon and the sun, it would be the difference between the distance they give us in the statement, because they have a point in common which is the earth, it would be:

1.496 x 10 ^ 8 km - 3.84 x 10 ^ 5 km = 149216000 = 1.49216 * 10 ^ 8 km

Therefore the distance between the moon and the sun is 1,49216 * 10 ^ 8 km

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Mofor’s school is selling tickets to the annual dance competition. On the first day of ticket sales the school sold 7 adult tick
zhannawk [14.2K]

The price of one adult ticket is $ 11 and the price of one student ticket is $ 11

<h3><u>Solution:</u></h3>

Given that , Mofor’s school is selling tickets to the annual dance competition.  

Let the cost of one adult ticket be $m and the cost of one student tickets be $n.

On the first day of ticket sales the school sold 7 adult tickets and 6 tickets for a total of $143.  

\text { Then, } 7 \times \text { cost of one adult ticket }+6 \times \text { cost of one student ticket }=\$ 143

7m + 6n = 143 ------- eqn (1)

The school took in $187 on the second day by selling 4 adult tickets and 13 student tickets.  

\text { Then, } 4 \times \text { cost of one adult ticket}+13 \times \text { cost of one student ticket}=\$ 187

4m + 13n = 187  ------ eqn (2)

We have to find the price of an adult ticket and the price of a student ticket.

Now, let us solve the equations.

Multiply eqn 1 by 4

28m + 24n = 572  ----- eqn 3

Multiply eqn 2 by 7

28m + 91n = 1309  ---- eqn 4

Now subtract eqn 4 from eqn 3

28m + 24n = 572

28m + 91n = 1309

(- )--------------------------------------

– 67n = - 737

67n = 737

n = 11

Plug in n = 11 in eqn 1

7m + 66 = 143

7m = 143 – 66

m = 11

Hence, the cost of one adult ticket is $ 11 and cost of one student ticket is $11

3 0
3 years ago
The product of 8, and a number increased by 7. is 104. What is the number?
Lynna [10]
Answer:
12.125
Step-by-step
8x+7=104
-7 -7
8x=97
/8 /8
x=12.125
4 0
3 years ago
Solve for the unknown side in the right triangle if b=24 and c=26 find a
cluponka [151]

Answer:

Option C

Step-by-step explanation:

Use the <u>Pythagorean Theorem</u> to find the unknown side.

a^2+b^2=c^2

<h3>We are given that:</h3>
  • b=24
  • c=26

We need to solve for a.

a^2+24^2=26^2\\\\ \text {Solve Exponents:}\\\rightarrow 24^2 = 576\\\rightarrow 26^2 = 676\\\\a^2+576=676\\\rightarrow a^2+576-576=676-576 \text{ (Subtraction Property of Equality)}\\a^2=100\\\sqrt{a^2} =\sqrt{100} \text { (Take the square root on both sides.) }\\\\\boxed {a=100}

<em>Option C should be the correct answer.</em>

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The answer for the question is 3/4
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3 years ago
which statement is always true A. the sum of two irrational numbers is rational. B. the product of a rational and irrational num
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The answer is D hope this helps
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