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3 years ago

6

While mining, Keith found a large metal bar that weighed 42 ounces. Keith was also able to determine that the bar had 30 ounces

of silver. What percent of the weight of the bar was silver? Round your answer to the nearest tenth if necessary.
PLEASE SHOW WORK!!!

2 answers:

Usimov [2.4K]3 years ago

8 0

Divide the silver present in the bar by the total weight of the bar to find the percentage of silver in the bar.

$30 / 42 = 0.7142$

Multiply this decimal by 100 for the percentage.

$.7142 * 100 = 71.42$

Rounded to the nearest tenth, 71.4% of the weight of the bar is silver.

Arada [10]3 years ago

6 0

70% is the answer......

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Irina-Kira [14]

In order to find the equation of a line, you would have to use the equation y=mx+b.

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Next in order to find b, you have to plug in a point:
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So your answer is: y=8x

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3 years ago

What is <br><img src="https://tex.z-dn.net/?f=8.4%20-%204.2" id="TexFormula1" title="8.4 - 4.2" alt="8.4 - 4.2" align="absmiddle

STALIN [3.7K]

Answer:

4.2

Step-by-step explanation:

$\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}$

$\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}$

$\begin{matrix}\space\space&8&.&4\\ -&4&.&2\end{matrix}$

$\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:4-2=2$

$\frac{\begin{matrix}\space\space&8&.&\textbf{4}\\ -&4&.&\textbf{2}\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\textbf{2}\end{matrix}}$

$\mathrm{Place\:the\:decimal\:point\:in\:the\:answer\:directly\:below\:the\:decimal\:points\:in\:the\:terms}$

$\frac{\begin{matrix}\space\space&8&\textbf{.}&4\\ -&4&\textbf{.}&2\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{.}&2\end{matrix}}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:8-4=4$

$\frac{\begin{matrix}\space\space&\textbf{8}&.&4\\ -&\textbf{4}&.&2\end{matrix}}{\begin{matrix}\space\space&\textbf{4}&.&2\end{matrix}}$

$=4.2$

3 0

2 years ago

151 241<br> 12 12<br> 102<br> 101<br> 35<br> Perimeter<br> ?<br> Simplify your answer completely

Nataliya [291]

9514 1404 393

Answer:

119 4/7

Step-by-step explanation:

The perimeter is the sum of side lengths.

10 1/7 + 12 2/7 + 15 3/14 + 24 1/14 + 12 2/7 + 10 1/7 + 35 3/7

= (10 + 12 + 15 + 24 + 12 + 10 + 35) + (1 + 2 + 2 + 1 + 3)/7 + (3 + 1)/14

= 118 + 9/7 + 4/14

= 118 + 1 2/7 + 2/7

= 119 4/7 . . . . perimeter of the figure

8 0

3 years ago

A rectangular box is to have a square base and a volume of 12 ft3. If the material for the base costs $0.17/ft2, the material fo

katen-ka-za [31]

Answer:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

Step-by-step explanation:

Let the dimensions of the box be x, y and z

The rectangular box has a square base.

Therefore, Volume of the box $V=x^2z$

Volume of the box $=12 ft^3\\$

$Therefore, x^2z=12\\z=\frac{12}{x^2}$

The material for the base costs $\$0.17/ft^2$ , the material for the sides costs $\$0.10/ft^2$ , and the material for the top costs $\$0.13/ft^2$ .

Area of the base $=x^2$

Cost of the Base $=\$0.17x^2$

Area of the sides $=4xz$

Cost of the sides= $=\$0.10(4xz)$

Area of the Top $=x^2$

Cost of the Base $=\$0.13x^2$

Total Cost, $C(x,z) =0.17x^2+0.13x^2+0.10(4xz)$

Substituting $z=\frac{12}{x^2}$

$C(x) =0.17x^2+0.13x^2+0.10(4x)(\frac{12}{x^2})\\C(x)=0.3x^2+\frac{4.8}{x} \\C(x)=\dfrac{0.3x^3+4.8}{x}$

To minimize C(x), we solve for the derivative and obtain its critical point

$C'(x)=\dfrac{0.6x^3-4.8}{x^2}\\Setting \:C'(x)=0\\0.6x^3-4.8=0\\0.6x^3=4.8\\x^3=4.8\div 0.6\\x^3=8\\x=\sqrt[3]{8}=2$

Recall: $z=\frac{12}{x^2}=\frac{12}{2^2}=3\\$

Therefore, the dimensions that minimizes the cost of the box are:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

7 0

3 years ago

Twelve of the 40 cars pulled by the locomotive were tankers.

quester [9]

12 are tankesr
40 cars total
12/40
12/40=6/20=3/10

3/10=tankders
10/10-3/10=7/10
7/10=70/100=70%
70% wern't tankser

3 0

3 years ago