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Tpy6a [65]
3 years ago
12

Isaiah and his children went into a grocery store and where they sell apples for $0.50 each and peaches for $1.50 each. Isaiah h

as $15 to spend and must buy no less than 14 apples and peaches altogether. If Isaiah decided to buy 8 apples, determine the maximum number of peaches that he could buy. If there are no possible solutions, submit an empty answer.
Mathematics
2 answers:
pychu [463]3 years ago
8 0

Answer:

8x0.50=4

15-4=9

9/1.50= 6

6 peaches

erica [24]3 years ago
6 0

Answer:

Maximum of <u>6 peaches</u>, and 8 apples

Explanation:

Given that the quantity of apples and peaches must total to 14, a total cost of $15, with a cost of $0.50 each for apples and a cost of $1.50 each for peaches, we can create a system of equations to solve for each variable.

0.5a + 1.50p = 15.

a + p = 14.

There are many methods that can be used but the easiest one that works for this problem is elimination.

Start by multiplying the top equation by 2 to get rid of the decimals.

0.5 a × 2 + 1.5 p × 2 = 15 × 2.

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Can someone explain the steps to solve number two?
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Ok so first you would create a proportion.
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Read 2 more answers
Greg is trying to solve a puzzle where he has to figure out two numbers, x and y. Three less than two-third of x is greater than
Fittoniya [83]
Inequation 1: 

\frac{2}{3}x-3 \geq y

to plot the pairs (x, y) for which the inequation holds, draw the line y=\frac{2}{3}x-3

then pick a point in either side of the line. If that point is a solution of the inequation, than color that region of the line, if that point is not a solution, then color the other part of the line.

we do the same for the second inequation. Then the solution, is the region of the x-y axes colored in both cases.

inequation 2: 

y+ \frac{2}{3}x\ \textless \ 4

y\ \textless \ - \frac{2}{3} x+ 4&#10;


draw the lines 

i)  y=\frac{2}{3}x-3          use points (0, -3),  (3, -1)

ii)y=- \frac{2}{3} x+ 4       use points ( 0, 4),   (3, 2)


let's use the point P(3, 3) to see what region of the lines need to be coloured:

\frac{2}{3}x-3 \geq y  ; 
\frac{2}{3}(3)-3 \geq 3
2-3 \geq 3, not true so we color the region not containing this point


y+ \frac{2}{3}x\ \textless \ 4
(3)+ \frac{2}{3}(3)\ \textless \ 4
3+ 5\ \textless \ 4 not true, so we color the region not containing the point (3, 3)

The graph representing the system of inequalities is the region colored both red and blue, with the blue line not dashed, and the red line dashed.



4 0
3 years ago
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