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Alla [95]
3 years ago
12

I need help Please help.

Mathematics
1 answer:
asambeis [7]3 years ago
8 0

Answer:

Step-by-step explanation:

Number of students 10

Problem 1. $625 for the bus hire per friday, So 625*4=$2500

Problem 2. 2500/25=$100 each for the whole 4 weeks

Problem 3.

10 students tickets 220= 2200 for all tickets. The bus, 625/10 = $62.5*4= $250 dollars for the whole 4 weeks for the bus so in all each student pays $470 each

20 students, tickets 220=4400 for all tickets. The bus, 625/20=$31.25*4=$125 for the whole 4 weeks for the bus, so in all each student must pay $345 each

30 students, tickets 220 = 6600 for all tickets. The bus, 625/30 =$20.83*4=$83.32 for the whole 4 weeks for the bus, so in all each student must pay $303.32 each

41 students, tickets $160=$6560 for all tickets. The bus, because you need 2 buses at 625 each so $1250 for both buses 1250/41= 30.49*4=$121.96 for the whole 4 weeks for the bus. So in al each student must pay $281.96 each

Hope this is correct

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A cafeteria sells 6 times as many bottles of juice. if the cafeteria sells 750 bottles of water. how many bottles of juice did t
faltersainse [42]

Answer:

<h2>4,500 bottles of juice.</h2>

Step-by-step explanation:

<h3>Clue word: Times means multiply</h3><h3>If you multiply 750 x 6 you will get 4,500.</h3>
5 0
2 years ago
X-y=5 and x^2y=5x+6​
sergeinik [125]

By applying algebraic handling on the two equations, we find the following three <em>solution</em> pairs: x₁ ≈ 5.693 ,y₁ ≈ 10.693; x₂ ≈ 1.430, y₂ ≈ 6.430; x₃ ≈ - 0.737, y₃ ≈ 4.263.

<h3>How to solve a system of equations</h3>

In this question we have a system formed by a <em>linear</em> equation and a <em>non-linear</em> equation, both with no <em>trascendent</em> elements and whose solution can be found easily by algebraic handling:

x - y = 5      (1)

x² · y = 5 · x + 6       (2)

By (1):

y = x + 5

By substituting on (2):

x² · (x + 5) = 5 · x + 6

x³ + 5 · x² - 5 · x - 6 = 0

(x + 5.693) · (x - 1.430) · (x + 0.737) = 0

There are three solutions: x₁ ≈ 5.693, x₂ ≈ 1.430, x₃ ≈ - 0.737

And the y-values are found by evaluating on (1):

y = x + 5

x₁ ≈ 5.693

y₁ ≈ 10.693

x₂ ≈ 1.430

y₂ ≈ 6.430

x₃ ≈ - 0.737

y₃ ≈ 4.263

By applying algebraic handling on the two equations, we find the following three <em>solution</em> pairs: x₁ ≈ 5.693 ,y₁ ≈ 10.693; x₂ ≈ 1.430, y₂ ≈ 6.430; x₃ ≈ - 0.737, y₃ ≈ 4.263.

To learn more on nonlinear equations: brainly.com/question/20242917

#SPJ1

8 0
2 years ago
If 5x:(x-2) = 11:3 , find the value of x ?​
Masteriza [31]

Answer:

x = -11/2

Step-by-step explanation:

The ratios need to stay the same

5x        11

----- = -------

x-2       3

Using cross products

5x*3 = (x-2) 11

15x = 11x - 22

Subtract 11 from each side

15x-11x = -22

4x = -22

x = -22/4

x = -11/2

5 0
2 years ago
Read 2 more answers
Nicole sold 1 cherry pie and 9 pumpkin pies for $60. Lisa sold 11 cheery pies and 4 pumpkin pies for $90. what's the cost each o
zubka84 [21]
X= cost per cherry pie
y= cost per pumpkin pie

NICOLE
1x + 9y= $60

LISA
11x + 4y= $90


STEP 1
multiply Nicole's equation by -11

-11(1x + 9y)= -11($60)
multiply -11 by all terms

(-11 * x) + (-11 * 9y)= (-11 * 60)

-11x - 99y= -660


STEP 2
add Nicole's new equation from step 1 to Lisa's equation to solve for y (using the elimination method)

-11x - 99y= -660
11x + 4y= 90
the x terms "cancel out"

-95y= -570
divide both sides by -95

y= $6 per pumpkin pie


STEP 3
substitute y value into either original equation to solve for x

x + 9y= $60

x + 9(6)= 60

x + 54= 60
subtract 54 from both sides

x= $6 per cherry pie


CHECK
11x + 4y= $90
11(6) + 4(6)= 90
66 + 24= 90
90= 90


ANSWER: Each cherry pie costs $6 and each pumpkin pie costs $6.

Hope this helps! :)
3 0
3 years ago
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levacccp [35]
X=11ft
25-6=14
So ur answer is actually 3
3 0
3 years ago
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