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Firdavs [7]
3 years ago
13

5.5r - 8.35s when r = 12 and s = 4.

Mathematics
1 answer:
nataly862011 [7]3 years ago
6 0

Answer:

32.6

Step-by-step explanation:

just do this

5.5(12)-8.35(4)

then solve

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What is 2,069 is in the written
777dan777 [17]

Answer:

2.069x10^3

Step-by-step explanation:

Move the decimal places.

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3 years ago
A candy store owner wants to mix some candy costing $1.25 a pound
maria [59]

The candy store owner should use 37.5 pounds of the candy costing $1.25 a pound.

Given:

  • Candy costing $1.25 a pound is to be mixed with candy costing $1.45 a pound
  • The resulting mixture should be 50 pounds of candy
  • The resulting mixture should cost $1.30 a pound

To find: The amount of candy costing $1.25 a pound that should be mixed

Let us assume that the resulting mixture should be made by mixing 'x' pounds of candy costing $1.25 a pound.

Since the total weight of the resulting mixture should be 50 pounds, 'x' pounds of candy costing $1.25 a pound should be mixed with '50-x' pounds of candy costing $1.45 a pound.

Then, the resulting mixture contains 'x' pounds of candy costing $1.25 a pound and '50-x' pounds of candy costing $1.45 a pound.

Accordingly, the total cost of the resulting mixture is 1.25x+1.45(50-x)

However, the resulting mixture should be 50 pounds and should cost $1.30 a pound. Accordingly, the total cost of the resulting mixture is 1.30 \times 50

Equating the total cost of the resulting mixture obtained in two ways, we get,

1.25x+1.45(50-x)=1.30 \times 50

1.25x+72.5-1.45x=65

0.2x=7.5

x=\frac{7.5}{0.2}

x=37.5

This implies that the resulting mixture should be made by mixing 37.5 pounds of candy costing $1.25 a pound.

Learn more about cost of mixtures here:

brainly.com/question/17109505

4 0
3 years ago
Read 2 more answers
Winning the jackpot in a particular lottery requires that you select the correct four numbers between 1 and 42 ​and, in a separa
NARA [144]

Answer:

\dfrac{1}{3,357,900}

Step-by-step explanation:

There are

C^{42}_4=\dfrac{42!}{4!(42-4)!}=\dfrac{42!}{4!\cdot 38!}=\dfrac{38!\cdot 39\cdot 40\cdot 41\cdot 42}{2\cdot 3\cdot 4\cdot 38!}=39\cdot 10\cdot 41\cdot 7=111,930

different ways to select the  four numbers between 1 and 42. Only one of this ways is correct (successful to win).

There are 30 different ways select the single number between 1 and 30. Only one of them is correct.

The  probability of winning the jackpot is

\dfrac{1}{111,930}\cdot \dfrac{1}{30}=\dfrac{1}{3,357,900}

4 0
3 years ago
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Eric wants to attend a public four-year university. He estimates that the average cost of one year at this university is $19,500
rjkz [21]
325 * 60 = 19500 so 60 months
7 0
3 years ago
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Two vertical poles of length 10 and 12 ​feet, respectively, stand 17 feet apart. A cable reaches from the top of one pole to the
Marat540 [252]

Answer:

L=\sqrt{x^{2}+100}+\sqrt{x^{2}-34x+433} ft

Step-by-step explanation:

let length of calble = L

distance from 10-foot pole = x ft

Moving from the top of the 10-foot pole to point x, we have

From the diagram that

x^{2} +10^{2}=L_{1} ^{2}  \\x^{2} + 100 = L_{1} ^{2}

Take the square root of both sides

\sqrt{x^{2}+100}=\sqrt{L_{1} ^{2}}\\  \sqrt{x^{2}+100}= L_{1}

Moving from point x to the top of the 12-foot pole, we have

From the diagram that

(17-x)^{2}+12^{2}=L_{2} ^{2}   \\289-34x+x^{2} +144=L_{2} ^{2}\\x^{2} -34x+433=L_{2} ^{2}

Take the square root of both sides

\sqrt{x^{2}-34x+433}=\sqrt{L_{2} ^{2}}  \\\sqrt{x^{2}-34x+433}=L_{2}

Amount of cable used, L = L1 + L2

L=\sqrt{x^{2}+100}+\sqrt{x^{2}-34x+433}

3 0
3 years ago
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