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2 years ago

Pre college need helppp

Mathematics

1 answer:

Brrunno [24]2 years ago

7 0

Answer:

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Find the missing measure in the triangle below.

denis23 [38]

Answer:

Step-by-step explanation:

To find x take 180 and subtract 63 and 77

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3 years ago

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What is the equation of the line that passes through the points (9, 2) and (7, -4)?

aliya0001 [1]

Answer:m=3

Step-by-step explanation:

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3 years ago

Sameera predicted that she would sell 38 blankets, but she actually sold 28 blankets. Which expression would find the percent er

antoniya [11.8K]

Solution: We are given:

Predicted Sales by Sameera $=38$

Actual Sales by Sameera $=28$

Now to find the Percent error, we have to use the below formula:

$Percent-Error= \frac{|Predicted-value - Actual-value|}{Actual-value} \times 100 \%$

$=\frac{|38-28|}{28} \times 100 \%$

$=\frac{10}{28}\times 100 \%$

$=35.71 \%$

Therefore, the percent error is $35.71 \%$

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3 years ago

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Help me with this question please?

UNO [17]

$\bf 3^{n+2}-3^n=8\cdot 3^n\implies \boxed{3^n\cdot 3^2}-3^n=8\cdot 3^n
\\\\\\
\stackrel{common~factor}{3^n}(3^2-1)=8\cdot 3^n\implies 3^2-1=\cfrac{8\cdot \underline{3^n}}{\underline{3^n}}\implies 3^2-1=8
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9-1=8\implies 8=8$

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2 years ago

A metallurgist has one alloy containing 34% copper another containing 48% copper. How many pounds of each alloy must he use to m

Zielflug [23.3K]

Answer:

<h2>36.14 pounds of 34% copper alloy and 9.86 pounds of 48% copper alloy</h2>

Step-by-step explanation:

First alloy contains 34% copper and the second alloy contains 48% alloy.

We wish to make 46 pounds of a third alloy containing 37% copper.

Let the weight of first alloy used be $x$ in pounds and the weight of second alloy used be $y$ in pounds.

Total weight = $46\text{ }pounds=x+y$ $-(i)$

Total weight of copper = $37\%\text{ of 46 pounds = }34\%\text{ of }x\text{ pounds + }48\%\text{ of }y\text{ pounds }$

$\dfrac{37\times 46}{100}=\dfrac{34x}{100}+\dfrac{48y}{100}\\\\ 34x+48y=1702$ $-(ii)$

Subtracting 34 times first equation from second equation,

$34x+48y-34x-34y=1702-34\times46\\14y=138\\y=9.857\text{ }pounds \\x=36.143\text{ }pounds$

∴ 36.14 pounds of first alloy and 9.86 pounds of second alloy were used.

5 0

3 years ago