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alina1380 [7]
3 years ago
15

Suzy buys 2 shirts and 2 pairs of pants for 42 dollars at the store. Annette buys 3 shirts and 1 pair of pants for 37 dollars.

Mathematics
1 answer:
balandron [24]3 years ago
6 0

Answer:

Shirts:  $8

Pants:  $13

Step-by-step explanation:

2 shirts and 2 pairs of pants costs $42.

2x + 2y = 42

3 shirts and 1 pair of pants costs $37.

3x + y = 37

Use substitution to solve.  Rearrange the second equation so that it is equal to y.  Plug in the value for y into the first equation.

3x + y = 37

y = 37 - 3x

2x + 2y = 42

2x + 2(37 - 3x) = 42

2x + 74 - 6x = 42

74 - 4x = 42

-4x = -32

x = 8

Now that you have the value for x, solve for y using one of the equations.  I will use the first equation, but you can use the other one and still get the right answer.

2x + 2y = 42

2(8) + 2y = 42

16 + 2y = 42

2y = 26

y = 13

The shirts cost $8, and the pants cost $13.

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Answer:

x=4, x=3

B is correct.

Step-by-step explanation:

Given: x^2-7x+12=0

Using middle term splitting factor the left side equation.

x^2-4x-3x+12=0

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The non-algebraic functions are called transcendental functions. This include the logarithmic function. The definition of Logarithmic Function with Base a is as follows:


For \ x\ \textgreater \ 0, \ a \ \textgreater \ 0, \ and \ a \neq 1 \\ \\ y=log_{a}x \ if \ and \ only \ if \ x=a^{y} \\ \\ Then: \\ \\ f(x)=log_{a}x \\ \\ is \ called \ the \ logarithmic \ function \ with \ base \ a.


We know that the equations is:


y=log(10x)


So let's solve each case:


Case 1


x=\frac{1}{100} \\ \\ y=log(10(\frac{1}{100})) \\ \\ \therefore y=log(\frac{1}{10}) \\ \\ \therefore y=-1 \\ \\ So: \\ \\ \boxed{\ x=\frac{1}{100}} \ matches \ to \ \boxed{y=-1}


Case 2


x=\frac{1}{10} \\ \\ y=log(10(\frac{1}{10})) \\ \\ \therefore y=log(1) \\ \\ \therefore y=0 \\ \\ So: \\ \\ \boxed{\ x=\frac{1}{10}} \ matches \ to \ \boxed{y=0}


Case 3


x=1 \\ \\ y=log(10(1)) \\ \\ \therefore y=log(10) \\ \\ \therefore y=1 \\ \\ So: \\ \\ \boxed{\ x=1} \ matches \ to \ \boxed{y=1}


Case 4


x=10 \\ \\ y=log(10(10)) \\ \\ \therefore y=log(100) \\ \\ \therefore y=2 \\ \\ So: \\ \\ \boxed{\ x=10} \ matches \ to \ \boxed{y=2}


Case 5


x=100 \\ \\ y=log(10(100)) \\ \\ \therefore y=log(1000) \\ \\ \therefore y=3 \\ \\ So: \\ \\ \boxed{x=100} \ matches \ to \ \boxed{y=3}

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