Can 1 be an additive identity?

1 answer:

5 0

Hey there, Valensolar! I would usually explain my answer, but in this case, I cant use any of the symbols required. But the answer is no. The number 1 can NOT be an additive identity. I hope this helps! :D

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Pierre wants to make tables and chairs. He has a total of 220 wooden boards and 760 nails. 13T+8C≤220 represents the number of t

Alisiya [41]

Answer:

Pierre has enough boards and nails to make 9 tables and 5 chairs.

Step-by-step explanation:

13T+8C ≤ 220

Let us substitute 9 for T and 5 for C and check.

13(9) + 8(5) = 117 + 40

= 157 < 220

So, 220 wooden boards are sufficient to make 9 tables and 5 chairs.

48T+37C ≤ 760

Substitute 9 for T and 5 for C.

48(9)+37(5) = 432 + 185

= 617 < 760

So, 760 nails are sufficient to make 9 tables and 5 chairs.

Hence, Pierre has enough boards and nails to make 9 tables and 5 chairs.

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Vikentia [17]

Answer:

8% of 800 is 64

800-64=736

Step-by-step explanation:

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3 years ago

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sesenic [268]

keeping in mind that perpendicular lines have negative reciprocal slopes, hmmm what's the slope of the equation above anyway?

$\bf y = \cfrac{2}{3}x\implies y = \stackrel{\stackrel{m}{\downarrow }}{\cfrac{2}{3}}x+0\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{2}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{3}{2}}\qquad \stackrel{negative~reciprocal}{-\cfrac{3}{2}}}$

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$\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{0})~\hspace{10em} \stackrel{slope}{m}\implies -\cfrac{3}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{-\cfrac{3}{2}}(x-\stackrel{x_1}{0})\implies y=-\cfrac{3}{2}x$