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Alexeev081 [22]
3 years ago
9

CAN SOMEONE PLEASE HELP ME I HAVE POINTS AND BRAIBLEST !!

Mathematics
1 answer:
Vladimir79 [104]3 years ago
6 0
Answer:560; 0.4*1400=560
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Change the following to algebraic expressions<br><br>nine times a number decreased by 11​
Nitella [24]
<h2>X^9 - 11</h2>

Step-by-step explanation:

let the number be X

so,

X^9 - 11

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How do I get the answer for this quadratic equation in simplest form?
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Answer:

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x=7-√15

Step-by-step explanation:

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For the upcoming semester, I have a teaching load of 13hours, meaning that I am expected to be in the classroom for 13 hours a w
Burka [1]
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3 years ago
You decide to join a tennis class. The initial entrance fee is
qwelly [4]

Given :

Initial entrance fee of tennis class, E = $100 .

Fee of every month, F = $10 .

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How much will it cost for five months.

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Let, number of months are n and cost of class is y.

So, linear equation is given by :

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To know the cost of 5 month, putting value of n = 5 in equation 1) we get :

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3 years ago
Write an equation of a line that is perpendicular to the line y=2/3x and passes through origin
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}}}

so we're really looking for the equation of a line whose slope is -3/2 and runs through (0,0).

\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

7 0
3 years ago
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