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3 years ago

Secondary school falls between:

Business

2 answers:

Trava [24]3 years ago

8 0

Answer:

Secondary school is from grades 7-12 (middle school except for 6th grade and then all of high school)

So it falls between primary school and college

OleMash [197]3 years ago

5 0

Answer:

(Apex) elementary school and post-secondary school

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Some states don't allow teacher aides to A. take attendance. B. monitor the halls. C. supervise a bus. D. aid in instructing a c

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4 years ago

The laws passed by Congress and by state legislatures are called:

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3 years ago

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The three broad categories of message strategies include ______________.

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4 years ago

A metallurgist has one alloy containing 26%26% copper and another containing 69%69% copper. How many pounds of each alloy must h

Brilliant_brown [7]

Answer:

Ans. He must use 2,987.76 pounds of alloy X (Cu=69%) and 2,365.24 pounds of alloy Y (26%=Cu)

Explanation:

Hi, let´s call alloy X the alloy that contains 69% of Cu and alloy Y the one containing 26% of Cu. Since he needs to produce 5,353 pounds of alloy, the first equation that we need to use is the following.

$X+Y=5,353$

Now, we need this 5,353 pounds of the new alloy to be 50% Cu, therefore, we have to use a portion of alloy X and alloy Y. This is as follows.

$0.69X+0.26Y=0.5(X+Y)$

And we have already established that X+Y is equal to 5,353, therefore this equation should look like this.

$0.69X+0.26Y=5,353*0.5$

$0.69X+0.26Y=2,676.5$

And we solve for X this equation, this as follows.

$0.69X=2,676.5-0.26Y$

$X=\frac{2,676.5-0.26Y}{0.69}$

$X=3,878.98-0.3768Y$

Now, we use this result and substitute this for X in the first equation like this.

$3,878.98-0.3768Y+Y=5,353$

and then, we solve for Y

$0.6232Y=5,353-3,878.98$

$Y=\frac{1,474.02}{0.6232} =2,365.24$

So, he needs to use 2,365.24 pounds of alloy that contains 26% of Cu, this means that the rest (2,987.76 pounds) must come from the alloy that contains 69% of Cu.

We can check this result by finding the overall Cu obtained by this amounts of alloy, that is

$Coppper FromX=2,987.76*0.69=2,061.55$

$Coppper FromY=2,365.24*0.26=614.96$

That adds up to 2,676.51 pounds of pure Cu, and since the total weight of the new alloy is 5,353, this amount of copper makes up for:

$PercentCu=\frac{2,676.51}{5,353} =0.5$

50% of the total weight.

Best of luck.

7 0

3 years ago