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yan [13]
3 years ago
12

HURRYYYYY !!!!! Examine the following expression. p squared minus 3 + 3 p minus 8 + p + p cubed Which statements about the expre

ssion are true? Check all that apply. The constants, –3 and –8, are like terms. The terms 3 p and p are like terms. The terms in the expression are p squared, negative 3, 3 p, negative 8, p, p cubed. The terms p squared, 3 p, p, and p cubed have variables, so they are like terms. The expression contains six terms. The terms p squared and p cubed are like terms. Like terms have the same variables raised to the same powers.
Mathematics
1 answer:
algol [13]3 years ago
6 0

Answer:

The terms 3p and p are like terms.

The terms in the expression are p squared, negative 3, 3p, p, negative 8, p, p cubed.

The expression contains 6 terms.

The terms p squared and p cubed are like terms.

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keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

\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}\qquad \qquad y = \stackrel{\stackrel{m}{\downarrow }}{3}x-5

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(\stackrel{x_1}{1}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \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}{(-3)}=\stackrel{m}{-\cfrac{1}{3}}(x-\stackrel{x_1}{1})\implies y+3=-\cfrac{1}{3}x+\cfrac{1}{3} \\\\\\ y=-\cfrac{1}{3}x+\cfrac{1}{3}-3\implies y=-\cfrac{1}{3}x-\cfrac{8}{3}

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2 years ago
Can you help me solve this radical equation √y-10 = y-2 the sqrt sign is only over the y-10
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D:y-10\geq0 \wedge y-2\geq0\\
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