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Korvikt [17]
3 years ago
13

48oz sprite for $1.39 or 32 oz dr. pepper for $0.89?

Mathematics
1 answer:
leonid [27]3 years ago
7 0
Answer: Dr. Pepper


Divide 1.39 by 48. Then divide 0.89 by 32.
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Noah drew 22 hearts and 76 circles . what is the ratio of circles to all shapes?
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The answer is 76/98 because 22+76 is 98
3 0
3 years ago
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Hi can anybody explain to me how to do this I'm kinda confused . 5×[5(-7)]​
telo118 [61]

Answer:

5 × [5(-7)]

= 5 × (-35)   ---- 5 × (-7) = -35

= -175

4 0
3 years ago
Write the inequality for the graph
Svetach [21]

Answer:

x≤-2

Step-by-step explanation:

it would be that because if the graph is going to the left that means it is less than x and the opposite is for greater than. Now if there is a closed circle that means the sign is equal to hence the line under the sign. So therefore it's x≤-2

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I have 2 triangle faces. i have 3 rectanglar faces. i have 6 vertices. i have 9 edges. what am i?
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Answer:

You are a rectangular prism.

Step-by-step explanation:

Rectangular prisms fill the criteria you described: two triangular faces, three rectangular faces, 6 vertices, and 9 edges. Thus, the answer is correct.

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8 0
3 years ago
Find the equation of a line that is perpendicular to y = 3x – 5 and passes through the point (1, -3).
Svetradugi [14.3K]

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

well then therefore

\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}

so we're really looking for the equation of a line with slope of -1/3 and that passes through (1, -3 )

(\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}

6 0
2 years ago
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