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3 years ago

Simplify the expression and find the value, if x=2 : 5x-6-x+9

Mathematics

2 answers:

timurjin [86]3 years ago

6 0

Answer:11

Step-by-step explanation:

5x-6-x+9

When x=2 we have

5(2)-6-2+9

10-6-2+9

10+9-6-2

19-6-2=11

Bezzdna [24]3 years ago

4 0

Answer:

Step-by-step explanation:

5x - 6 - x + 9

4x + 3

x = 2

4(2) + 3 = 11

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A circle has a radius of 26.85 inches. What is the circumference of this circle? Use 3.14 for pi. Round your final answer to the

Ad libitum [116K]

Answer:

Step-by-step explanation:

Circumference of the circle = 2 x pi x r

= 2 x 3.14 x 26.85

= 168.618 Inches

after rounding to nearest tenth, 168.6 inches

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3 years ago

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Angel is making bags of trail mix for hiking club. He will use 20 ounces of walnuts, 11.2 ounces of almonds, and 14.3 ounces of

zheka24 [161]

Answer:

1.82 oz.

Step-by-step explanation:

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3 years ago

What is 5.316 - 1.942 (show ur work)

Fiesta28 [93]

Answer:

3.374

Step-by-step explanation:

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

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

$\begin{matrix}\:\:&5&.&3&1&6\\ -&1&.&9&4&2\end{matrix}$

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

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:6-2=4$

$\frac{\begin{matrix}\:\:&5&.&3&1&\textbf{6}\\ -&1&.&9&4&\textbf{2}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\textbf{4}\end{matrix}}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}$

$\frac{\begin{matrix}\:\:&5&.&3&\textbf{1}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\mathrm{The\:bottom\:number\:is\:larger\:than\:the\:upper\:number.\:\:Try\:to\:'borrow'\:a\:digit\:from\:the\:left.}$

$\mathrm{The\:top\:digit\:is\:not\:bigger\:than\:the\:bottom\:one.\:\:Try\:to\:'borrow'\:a\:digit\:from\:the\:left.}$

$\frac{\begin{matrix}\:\:&5&.&\textbf{3}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\mathrm{Borrow\:}1\mathrm{\:from\:}5\mathrm{.\:\:The\:remainder\:is\:}4$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&10&\:\:&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&3&1&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{\:\:}&\:\:&\:\:&\:\:&4\end{matrix}}$

$\mathrm{Add\:}1\mathrm{\:ten\:to\:}3:\quad \:10+3=13$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{13}&\:\:&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{3}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\mathrm{Borrow\:}1\mathrm{\:from\:}13\mathrm{.\:\:The\:remainder\:is\:}12$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&10&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\mathrm{Add\:}1\mathrm{\:ten\:to\:}1:\quad \:10+1=11$

$\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:11-4=7$

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

$\frac{\begin{matrix}\:\:&4&\textbf{\:\:}&12&11&\:\:\\ \:\:&\linethrough{5}&\textbf{.}&\linethrough{13}&\linethrough{1}&6\\ -&1&\textbf{.}&9&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\textbf{.}&3&7&4\end{matrix}}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:4-1=3$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&12&11&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&\linethrough{13}&\linethrough{1}&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{3}&.&3&7&4\end{matrix}}$