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Reil [10]
3 years ago
7

How much pasta will be used

Mathematics
2 answers:
lutik1710 [3]3 years ago
7 0
I hope this helps you

Lesechka [4]3 years ago
5 0
C. 10lb. If one pound feeds 6 people than you would need to divide 60 by 6. 60÷6=10
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The equation y=0.5x+40 represents the monthly cost y in dollars of Lesley's cell phone, where x is the of talk minutes over 750
Firdavs [7]
You already have the equation given which is <span>y=0.5x+40. To graph, you just have to replace random values of x to determine the corresponding values of y. Plot these points and connect them. The graph is shown in the attached picture. As you can observe, the range starts from y=40. This is because the y-intercept is 40. So, you don't have to show the y-values below because it would just minimize your linear graph.</span>

7 0
3 years ago
A bike store sells scooters at a 54% markup if the store sells each scooter for $69.30 then what is their non-markup price
liq [111]

Answer:

$45

Step-by-step explanation:

1.54(x) = 69.3

Divide both sides by 1.54

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3 years ago
If a discriminant is negative, then the quadratic has:
Serjik [45]

Step-by-step explanation:

the answer is

........b

7 0
3 years ago
Read 2 more answers
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
Determine the approximate measure of Y.
Soloha48 [4]

Answer:

I took the test, Its B) 41.1

7 0
3 years ago
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