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EastWind [94]
2 years ago
9

What’s the answer plzzzzzzzz?

Mathematics
2 answers:
maks197457 [2]2 years ago
7 0

Step-by-step explanation:

please forgive me if I have done something wrong there I am in a hurry I have to go ccooking. if there's something wrong there you can tell me I check it out when I come back good luck.

Galina-37 [17]2 years ago
6 0

Answer:

1. 4310.3

2.1809.6

3. 2414.7

4. 230.9

5. 767.8

6. 70.3

7.1143.4

8.125.7

9. 1382.0

10. 158.5

Step-by-step explanation:

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the larger of two numbers is eight more than the smaller number. their sum is twenty-two. find the number
lakkis [162]
The bigger number is 15 and the smaller number 7. 15 is 8 more than 7. The sum of 15 and 7 is 22.
4 0
3 years ago
If 111 people attend a concert and tickets for adults cost $4 while tickets for children cost $3.25 and total receipts for the c
VikaD [51]

Answer:

57 children

54 adults

Step-by-step explanation:

Let's call x the number of children admitted and call z the number of adults admitted.

Then we know that:

x + z = 111

We also know that:

3.25x + 4z = 401.25

We want to find the value of x and z. Then we solve the system of equations:

-Multiplay the first equation by -4 and add it to the second equation:

-4x - 4z = -444

3.25x + 4z = 401.25

----------------------------------

-0.75x = -42.75

x =\frac{-42.75}{-0.75}\\\\x=57

Now we substitute the value of x in the first equation and solve for the variable z

57 + z = 111

z = 111-57

z = 54

5 0
2 years ago
Read 2 more answers
Prove that:
devlian [24]

Answer:

<u>Identities used:</u>

  • <em>1/cosθ = secθ</em>
  • <em>1/sinθ = cosecθ</em>
  • <em>sinθ/cosθ = tanθ</em>
  • <em>cosθ/sinθ = cotθ</em>
  • <em>sin²θ + cos²θ = 1</em>
<h3>Question 1 </h3>
  • (1 - sinθ)/(1 + sinθ) =        
  • (1 - sinθ)(1 - sinθ) / (1 - sinθ)(1 + sinθ) =
  • (1 - sinθ)² / (1 - sin²θ) =
  • (1 - sinθ)² / cos²θ

<u>Square root of it is:</u>

  • (1 - sinθ)/ cosθ =
  • 1/cosθ - sinθ / cosθ =
  • secθ - tanθ
<h3>Question 2 </h3>

<u>The first part without root:</u>

  • (1 + cosθ) / (1 - cosθ) =
  • (1 + cosθ)(1 + cosθ) / (1 - cosθ)(1 + cosθ)
  • (1 + cosθ)² / (1 - cos²θ) =
  • (1 + cosθ)² / sin²θ

<u>Its square root is:</u>

  • (1 + cosθ) / sinθ =
  • 1/sinθ + cosθ/sinθ =
  • cosecθ + cotθ

<u>The second part without root:</u>

  • (1 - cosθ) / (1 + cosθ) =
  • (1 - cosθ)²/ (1 + cosθ)(1 - cosθ) =
  • (1 - cosθ)²/ (1 - cos²θ) =
  • (1 - cosθ)²/sin²θ

<u>Its square root is:</u>

  • (1 - cosθ) / sinθ =
  • 1/sinθ - cosθ / sinθ =
  • cosecθ - cotθ

<u>Sum of the results:</u>

  • cosecθ + cotθ + cosecθ - cotθ =
  • 2cosecθ
4 0
2 years ago
Read 2 more answers
Oliver and Lenny started hiking a hill simultaneously from the bottom to the top. Oliver hiked at a speed of 5 km/h during the f
Papessa [141]

Answer:

<u>Oliver and Lenny reached the top of the hill at the same time</u>

<u>The proportion of time Oliver spent hiking to that of Lenny is 1:1</u>

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Oliver's speed = 5 km/h during the first half, then decreased his speed to 4 km/h during the second half.

Lenny's speed = 4 km/h during the first half, then increased his speed to 5 km/h in the second half.

2. Who reached the top of the hill first? What is the proportion of time he spent hiking to that of his friend?

Let's assume the distance from the bottom to the top of the hill is 10 kilometers for calculating the time it will take Oliver and Lenny to reach the top of the hill:

Speed = Distance/Time

Time = Distance/Speed

Oliver = Time of hiking first 5 km + Time of hiking second 5 km

Lenny = Time of hiking first 5 km + Time of hiking second 5 km

Oliver = 5/5 + 5/4

Oliver = 1 + 5/4 = 2 1/4 hours

Lenny = 5/4 + 5/5

Lenny = 5/4 + 1 = 2 1/4 hours

<u>Oliver and Lenny reached the top of the hill at the same time</u>

<u>The proportion of time Oliver spent hiking to that of Lenny is 1:1</u>

3 0
2 years ago
Which number is a solution of the inequality? 6&gt;z(10-z)
I am Lyosha [343]
A.0 I think ................
6 0
3 years ago
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