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Maurinko [17]
3 years ago
6

Write an equation that involves

Mathematics
2 answers:
Romashka [77]3 years ago
7 0

so the solution is 5 we'll start with that

x=5

to do reverse multiplcation, you must divide so 

x=5

x/10=5/10

x/10=1/2

then the other problem

x=5

5x=25

to solvve

x/10=1/2

multiply by 10

x=5

then 

5x=25

divide both sides by 5

x=5

Lilit [14]3 years ago
6 0
We can use any equation, as long as it’s solution is 5.

So, 4x(5+x)-2(5x+2x^2)=50

10x=50

x=5

Now, to use division:

20x/10+ 4x^2/4x^2=11

Simplifying we get:

2x+1=11

2x=10
x=5.

This is a little complicated, but I hope this helps!
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Marcy purchases a total of 12 adult, child, and senior tickets for an art museum tour. Adult tickets are $5 each, child tickets
Kruka [31]

Answer:

A: The solution to this system is no viable because it results in fractional values of tickets

Step-by-step explanation:

Let number of adult, child, and senior tickets be x, y and z respectively.

Thus;

x + y + z = 12 - - - (eq 1)

she purchases 4 more child tickets than senior tickets.

Thus, y = z + 4

Thus:

x + (z + 4) + z = 12

x + 2z + 4 = 12

x + 2z = 8 - - - (eq 2)

Also, We are told that Adult tickets are $5 each, child tickets are $2 each, and senior tickets are $4 each. She spends a total of $38,

Thus;

5x + 2(z + 4) + 4z = 38

5x + 2z + 8 + 4z = 38

5x + 6z + 8 = 38

5x + 6z = 38 - 8

5x + 6z = 30

Divide through by 5 to get;

x + (6/5)z = 6 - - - (eq 3)

Subtract eq 2 from eq 1 to get;

2z - (6/5)z = 8 - 6

2z - (6/5)z = 2

Multiply through by 5 to get;

10z - 6z = 10

4z = 10

z = 2½

It's not possible to have a fractional value of a ticket and thus we can say that the solution is not viable.

6 0
3 years ago
A. True/False
ra1l [238]

Answer:

True

Step-by-step explanation:

x^{2} -6x=9

subtract 9 from both sides

x^{2} -6x-9

<em>plz mark me brainliest.</em> ;)

5 0
3 years ago
an open box is to be made from a piece of metal 16 by 30 inches by cutting out squares of equal size from the corners and bendin
blagie [28]
Let the lengths of the bottom of the box be x and y, and let the length of the squares being cu be z, then
V = xyz . . . (1)
2z + x = 16 => x = 16 - 2z . . . (2)
2z + y = 30 => y = 30 - 2z . . . (3)

Putting (2) and (3) into (1) gives:
V = (16 - 2z)(30 - 2z)z = z(480 - 32z - 60z + 4z^2) = z(480 - 92z + 4z^2) = 480z - 92z^2 + 4z^3
For maximum volume, dV/dz = 0
dV/dz = 480 - 184z + 12z^2 = 0
3z^2 - 46z + 120 = 0
z = 3 1/3 inches

Therefore, for maximum volume, a square of length 3 1/3 (3.33) inches should be cut out from each corner of the cardboard.
The maximum volume is 725 25/27 (725.9) cubic inches.
8 0
3 years ago
How many solutions does this linear system of equations have?
Irina18 [472]
It has infinite solutions
4 0
3 years ago
Read 2 more answers
What is nine plus ten?
poizon [28]
Twenty one

9 + 10

= 19
7 0
3 years ago
Read 2 more answers
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