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Keith_Richards [23]
3 years ago
15

The price of Philip's favorite candy has been reduced by $0.15 per piece. Philip buys 14 pieces and spends $3.22.

Mathematics
2 answers:
kompoz [17]3 years ago
6 0
3.22 divided by 14 is .23.   .23 plus .15 is .38 cents per candy at regular price.
dimulka [17.4K]3 years ago
5 0

For this case, the first thing we should do is define a variable.

We have then:

  • <em>x: regular price of each piece. </em>

The equation modeling the problem is given by:

14 (x-0.15) = 3.22

From here, we clear the value of x.

We have then:

x-0.15 = \frac {3.22} {14}\\x-0.15 = 0.23\\x = 0.23 0.15\\x = 0.38\$

Answer:

The regular price of each piece of candy is:

x = 0.38\$


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How do you find the <br> y-intercept and the slope for number 11
Sophie [7]
The first step is to find the slope. Use the slope formula

m = (y2-y1)/(x2-x1)

The two points are (x1,y1) = (-1,5) and (x2,y2) = (2,-1)

So,
x1 = -1
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and
x2 = 2
y2 = -1
will be plugged into the slope formula to get...
m = (y2-y1)/(x2-x1)
m = (-1-5)/(2-(-1))
m = (-1-5)/(2+1)
m = (-6)/(3)
m = -2

The slope is -2

Use m = -2 and one of the points to find the y intercept b. I'll use the point (x,y) = (-1,5) ---> x = -1, y = 5

y = mx+b ... slope intercept form
5 = -2*(-1)+b
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3 = b
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The y intercept is 3

-----------------------------------

m = -2 is the slope
b = 3 is the y intercept

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8 0
3 years ago
Evaluate log 2 base 8​
iren2701 [21]

Answer:  The correct answer is:  " 3 " .

_______________________________

Step-by-step explanation:

_______________________________

The problem given is:

_______________________________

Evaluate:  log 2 base 8 .

_______________________________

Rewrite as:  " log_2}8 = ? _______________  " ;  log_2}8 = ? _______________

_______________________

Take note of the following definition of a logarithm:

_______________________

"   log_{m}n = a " ;  ↔  " m^{a} =n  "  ;

_______________________

in which:  m  refers to the "base" ;

               n  refers to the "argument" ;

               a  refers to the "exponent" ;

______________________________

As such:

       " log_2}8 = ? _______________ "

      ⇔ " 2^{x} = 8 " ; Solve for "x" ;  

 _______________

 Note:  

  2⁰ = 1 ;

  2¹ = 2 ;

  2² = 2 * 2 = 4 ;

  2³ = 2 * 2 * 2 = 8 .

___________________

 → " 2^{x} = 8 " ;

 → " 2^{3} = 8 " ;  since: " 2^{3} = 2  *2 *2 = 8 " ;

 → as such:   " x = 3  " ;

______________________

Thus:

"log 2 base 8 " ;

          = log{_{2}}8 = 3  .

_______________________

Hope this helps!

Best wishes to you!

3 0
3 years ago
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