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astraxan [27]
3 years ago
15

HELPPPPPPP ASAPPPPPP PLEASEEEEEEE

Mathematics
1 answer:
ANTONII [103]3 years ago
5 0

Answer:

4x-6y=5

if it is correct answer then please follow me

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A ball is thrown from the top of the school. The height of the ball (in feet) written in
seropon [69]

Answer:

3 seconds

Step-by-step explanation:

When the ball hits the ground, the height will be 0. Therefore, you can substitute 0 in for the value of f(t), hence the equation will be:

0 = -16t^2 + 32t + 48

You can then factor out -16:

0 = t^2 - 2t - 3

Factor:

0 = (t-3)(t+1)
t = 3, -1

However -1 is an extraneous solution because time cannot be negative. Therefore, the answer is 3 seconds.

7 0
2 years ago
Write an expression using fractions to show how to determine the amount that each person will pay. Then calculate each person's
DiKsa [7]

The fraction based on the information given illustrated for each person.

<h3>How to illustrate the information?</h3>

Me: $14.89

Friend 1: $7.27

Friend 2: $25.67

Friend 3: $11.59

Total = $59.42

The fraction for each person will be:

Me: $14.89 = 14.89/59.42 = 0.25

Friend 1: $7.27 = 7.27/59.42 = 0.12

Friend 2: $25.67 = 25.67/59.42 = 0.32

Friend 3: $11.59 = 11.59/59.42 = 0.195

Learn more about fractions on:

brainly.com/question/78672

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7 0
2 years ago
Please help!!!!!!!!!!!!!!!!!
egoroff_w [7]

2 not given 3 given 4given 5 not given

5 0
3 years ago
.
dimulka [17.4K]

Answer:

Area = 5x^2 +7x -26

Step-by-step explanation:

The area of the shaded region can be found if you substruct the small rectangle from the big one. The area of any rectangle is calculated if you multiply width and height.

In other words:

A_small = (x-3)(x-6) = x^2-9x +18

A_big = (2x+2)(3x-4) = 6x^2 -2x -8

A_big - A_small = (6x^2 -2x -8) - (x^2-9x +18)

= 6x^2 -2x -8 - x^2 + 9x -18

= 5x^2 +7x -26

4 0
3 years ago
In how many distinct ways can the letters of the word mathematics be arranged? (first, does the order matter?)
TEA [102]
Our current list has 11!/2!11!/2! arrangements which we must divide into equivalence classes just as before, only this time the classes contain arrangements where only the two As are arranged, following this logic requires us to divide by arrangement of the 2 As giving (11!/2!)/2!=11!/(2!2)(11!/2!)/2!=11!/(2!2).

Repeating the process one last time for equivalence classes for arrangements of only T's leads us to divide the list once again by 2
4 0
3 years ago
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