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Serggg [28]
3 years ago
8

Leila received a $90 gift card for a coffee store. She used it in buying some coffee that cost $8.64 per pound. After buying the

coffee, she had $64.08 left on her card. How many pounds of coffee did she buy?
Mathematics
1 answer:
klasskru [66]3 years ago
3 0
She bought 5 pounds of coffee
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Can someone please help me ?
Snowcat [4.5K]

The first answer in Question A is Business Plan and the second answer is $3 less.

The first answer in Question B is 12 and the second answer is Residential Plan.

4 0
3 years ago
1) Easton went to a concert with some
tatuchka [14]
Subtract the 15 from the 133
you get 118
divide 29.50 from the 118
you get 4
4 people went to the concert
7 0
3 years ago
What is the answer to |x-4|<13
VARVARA [1.3K]

Answer:

<h2>The solution is -9 < x < 17.</h2>

Step-by-step explanation:

|x-4|<13.

The above equation means, whatever the actual value of x is, the value of (x - 4) must be greater than - 13 and less than 13.

Hence, -13 < x - 4 < 13 or, -9 < x < 17. The value of x will be in between -9 and 17. The value of x can not be -9 or 17.

7 0
3 years ago
What is the slope of the line that passes through the pair of points (-2.5,6.1) and (-2.5,3.1)
podryga [215]
(-2.5,6.1)(-2.5,3.1).....notice how ur x values are the same. This means that u have a vertical line which has an undefined slope.

IF ur y values would have been the same, u would have had a horizontal line which has a 0 slope.
5 0
3 years ago
X - 3y = 17<br> x = 2y + 13<br> What are the steps for solving these?
lbvjy [14]

Answer:

  (x, y) = (5, -4)

Step-by-step explanation:

The second equation gives you an expression for x that can be substituted into the first equation.

  (2y +13) -3y = 17 . . substitute for x in the first equation

  -y = 4 . . . . . . . . . . . subtract 13 and simplify

  y = -4

  x = 2(-4) +13 = 5 . . . use the second equation to find the value of x

The solution is (x, y) = (5, -4).

_____

<em>Additional comment</em>

When faced with a system of linear equations, the first step is to look at them and observe where the variable terms are, and any relationships between coefficients. Several options are generally open to you for solving the equations. Methods generally taught first are ...

  • substitution
  • elimination

<em>Substitution</em> is handy when one of the variables or variable terms can be written (easily) in terms of the other variable. <em>Elimination</em> is handy when some simple multiple of one of the equations can be added to the other equation to cancel one of the variable terms (eliminate it).

Other available methods include matrix methods, Cramer's rule, and graphing. Working knowledge of all of these methods will help you identify the one that will be easiest to use in any given situation.

The availability of calculators able to use these methods greatly simplifies the task of finding a solution. It is simply a matter of entering the equations in the appropriate form.

One of my favorites is the graphing calculator, which only requires you type in the given equations and click on the solution point to find its coordinates.

4 0
2 years ago
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