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jek_recluse
1 month ago
6

Delaney would like to make a 10 lb nut mixture that is 60% peanuts and and 40% almonds. She has several pounds of peanuts and se

veral pounds of a mixture that is 20% peanuts and 80% almonds. Let p represent the number of pounds of peanuts needed to make the new mixture, and let m represent the number of pounds of the 80% almond-20% peanut mixture.
(a) What is the system that models this solution?
(b) Which of the following is a solution to the system: 4 lb peanuts and 6 lb mixture; 5 lb peanuts and 5 lb mixture; 8 lb peanuts and 2 lb mixture? Show your work.
Mathematics
1 answer:
AnnZ [12.3K]1 month ago
7 0

Response:

If there are "p" pounds of peanuts and "m" pounds of a mixture containing '20% peanuts and 80% almonds', then we can formulate the following equations:

p + m = 10 -----------(1) and

4m/5 = 4 ------------(2)

The solution yields 5 lb peanuts and 5 lb mixture.

Detailed explanation:

In the mixture that Delaney desires to create, there will be

\frac {10 \times 60}{100} lb

= 6 lb of peanuts

Thus, there will be (10 - 6) lb

= 4 lb of almonds

If Delaney has "p" pounds of peanuts and "m" pounds of the '20% peanuts and 80% almonds' mixture, then based on the problem statement,

p + m = 10 -----------(1) and

4m/5 = 4 ------------(2)

From equation (2), we derive

m = 5 --------------(3)

From (1) and (3), we find that

p = (10 - 5) = 5

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2 months ago
Let P be a point not on the line L that passes through the points Q and R. The distance d from the point P to the line L is d =
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Answer:

The distance from the point (0,1,1) to the specified line is zero.

Step-by-step explanation:

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In order to calculate the distance from (0,1,1), we must remove t from the equations above, such that

y=5-2t=5-x\implies x+y=5=4+1=4+z-t=4+z-\frac{1}{2}x

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\frac{al+bm+cn}{\sqrt{l62+m^2+n^2}}=\frac{(3\times 0)+(2\times 1)+(-2)(1)}{\sqrt{3^2+2^2+(-2)^2}}=\frac{0+2-2}{\sqrt{17}}=0

The distance between the point (0,1,1) and (1) amounts to zero. Therefore, the point (0,1,1) is located on the line (1).

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2 months ago
In Jane’s calculus class at a large university, the final exam had a mean score of 75 and a standard deviation of 10. In Peter’s
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Detailed breakdown:

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they paid $13 for each pepperoni pizza and $11for each plain pizza the club bought total of 18 pizzas for the party and paint a
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Complete question;

A booster club financed pizzas for an end-of-season celebration. The cost was $13 for each pepperoni pizza and $11 for each plain pizza. They purchased a

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Organize the numbers into an equation that represents this scenario with x being the number of pepperoni pizzas and y being the number of plain pizzas purchased.

Answer:

13x + 11y = 212

Step-by-step explanation:

Two varieties of pizza were purchased: pepperoni and plain. The pepperoni variety cost $13 each while the plain one cost $11 each. In total, they bought 18 pizzas.

The overall expenditure on the pizzas was $212.

total number of pizzas = 18

total amount spent on pizzas = $212

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The cost of the pepperoni pizzas is derived by multiplying the number of pizzas by their individual price. The same calculation applies for the plain pizzas.

Therefore, by summing the costs of both pizza types, we arrive at the total price for the pizzas.

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4 0
3 months ago
Jeremy had 3/4 of a submarine sandwich and gave
PIT_PIT [12445]
The actual question asks, "<span>Jeremy possessed 3/4 of a submarine sandwich and shared 1/3 of that with his friend. How much of the sandwich did his friend obtain?"

Solution:
Calculating 1/3 of 3/4 of a submarine sandwich results in = 1/3 x 3/4 = 3/12 = 1/4
Therefore, Jeremy gave his friend 1/4 of the submarine sandwich. </span>
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