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2 years ago

Brass is an alloy made by melting and mixing copper and zinc. A metallurgist has two brass

Mathematics

1 answer:

Drupady [299]2 years ago

4 0

Answer:

x +y = 500
0.65x +0.95y = 0.75(500)
solution: (x, y) = (300, 200)

Step-by-step explanation:

A system of equations for the problem can be written using the two given relationships between quantities of brass alloys.

<h3>Setup</h3>

Let x and y represent the quantities in grams of the 65% and 90% alloys used, respectively. There are two relations given in the problem statement.

x + y = 500 . . . . . . quantity of new alloy needed

0.65x +0.90y = 0.75(500) . . . . . quantity of copper in the new alloy

These are the desired system of equations.

<h3>Solution</h3>

This problem does not ask for the solution, but it is easily found using substitution for x.

x = 500 -y

0.65(500 -y) +0.90y = 0.75(500)

(0.90 -0.65)y = 500(0.75 -0.65) . . . . . . subtract 0.65(500)

y = 500(0.10/0.25) = 200

x = 500 -200 = 300

300 grams of 65% copper and 200 grams of 90% copper are needed.

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3 years ago

Jim needs to mix a medication with water to make the appropriate solution. The medications says to mix 3 oz of medication with 5

skelet666 [1.2K]

Answer:

38 or 38.33 cups of water.

Step-by-step explanation:

You need 5 cups of water for every 3 ounces of medication. We can set up a proportion: #oz of medication/#cups of water= #oz of medication/# cups of water. We can plug in some numbers from the problem.

3/5=23/n. You can use any variable, but I'm using n for how many cups of water Jim will need for 23 ounces of medication.

We can cross multiply: 3*n=5*23.

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We divide both sides by 3.

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3 years ago

Determine which of the sets of vectors is linearly independent. A: The set where p1(t) = 1, p2(t) = t2, p3(t) = 3 + 3t B: The se

defon

Answer:

The set of vectors A and C are linearly independent.

Step-by-step explanation:

A set of vector is linearly independent if and only if the linear combination of these vector can only be equalised to zero only if all coefficients are zeroes. Let is evaluate each set algraically:

$p_{1}(t) = 1$ , $p_{2}(t)= t^{2}$ and $p_{3}(t) = 3 + 3\cdot t$ :

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (3 +3\cdot t) = 0$

$(\alpha_{1}+3\cdot \alpha_{3})\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot t = 0$

The following system of linear equations is obtained:

$\alpha_{1} + 3\cdot \alpha_{3} = 0$

$\alpha_{2} = 0$

$\alpha_{3} = 0$

Whose solution is $\alpha_{1} = \alpha_{2} = \alpha_{3} = 0$ , which means that the set of vectors is linearly independent.

$p_{1}(t) = t$ , $p_{2}(t) = t^{2}$ and $p_{3}(t) = 2\cdot t + 3\cdot t^{2}$

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot t + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (2\cdot t + 3\cdot t^{2})=0$

$(\alpha_{1}+2\cdot \alpha_{3})\cdot t + (\alpha_{2}+3\cdot \alpha_{3})\cdot t^{2} = 0$

The following system of linear equations is obtained:

$\alpha_{1}+2\cdot \alpha_{3} = 0$

$\alpha_{2}+3\cdot \alpha_{3} = 0$

Since the number of variables is greater than the number of equations, let suppose that $\alpha_{3} = k$ , where $k\in\mathbb{R}$ . Then, the following relationships are consequently found: