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Klio2033 [76]
3 years ago
11

At a fish market, a pound of salmon costs $12 and a pound of shrimp costs $18. Mrs. Donnelly wants to buy 2 1/3 pounds of salmon

and 3 1/2 pounds of shrimp. If she has $93, can she buy the amounts of salmon and shrimp she wants? If yes, how much money will she have left over? If no, how much more does she need?
Mathematics
1 answer:
Semmy [17]3 years ago
7 0

Answer:

Yes and she would have $2 left over because all the food costs $91.

Step-by-step explanation:

We should set up an equation so that everything makes sense. A calculator would be wise for this problem, but I am going to lay it out anyhow.

We should convert the mixed numbers to improper fractions to not be confused. 3 1/2 becomes 7/2 or 3.5 and 2 1/3 becomes 7/3.

From your wording, we can assume that the money Mrs. Donnelly has is either greater than or equal to $93.

Thus, we have 12*(7/3) + 18*(7/3) is equal to or greater than 93 (this is an inequality).

The math shows that the food costed 91 and so she only has $2 left.

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Step-by-step explanation:

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m\angle CAB=m\angle CBA=30^{\circ} (base angles of an isosceles triangle are congruent)

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3 0
2 years ago
Solve for x and then part b) find JL. Have no clue how to go about this
Igoryamba

9514 1404 393

Answer:

  • x = 4
  • JL = 16

Step-by-step explanation:

a) You go about this by using the given equation.

  Left + Right = Total

  JK + KL = JL

  (2x -3) + 11 = 4x

  8 = 2x . . . . . simplify, subtract 2x

  4 = x . . . . . . divide by 2

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b) JL = 4x = 4·4

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2 years ago
AS one of her chores Staci must vacum 1/2 of the house. If she has already vacumed 1/3 of her share tonight whatp part of the li
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3 years ago
Which ordered pair is a solution to the system of linear equations One-half x minus three-fourths y = StartFraction 11 Over 60 E
s344n2d4d5 [400]

Answer:

\left(\dfrac{2}{3},\dfrac{1}{5}\right)

Step-by-step explanation:

Given the system of two equations:

\left\{\begin{array}{l}\dfrac{1}{2}x-\dfrac{3}{4}y=\dfrac{11}{60}\\ \\\dfrac{2}{5}x+\dfrac{1}{6}y=\dfrac{3}{10}\end{array}\right.

Multiply the first equation and the second equation by 60 to get rid of fractions:

\left\{\begin{array}{l}30x-45y=11\\ \\24x+10y=18\end{array}\right.

Now multiply the first equation by 4 and the second equation by 5:

\left\{\begin{array}{l}120x-180y=44\\ \\120x+50y=90\end{array}\right.

Subtract them:

(120x-180y)-(120x+50y)=44-90\\ \\120x-180y-120x-50y=-46\\ \\-230y=-46\\ \\y=\dfrac{46}{230}=\dfrac{1}{5}

Substitute it into the first equation:

30x-45\cdot \dfrac{1}{5}=11\\ \\30x-9=11\\ \\30x=11+9\\ \\30x=20\\ \\x=\dfrac{2}{3}

The solution is \left(\dfrac{2}{3},\dfrac{1}{5}\right)

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