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kupik [55]
3 years ago
12

Solve the inequality 3k > 5k+12

Mathematics
1 answer:
Natali5045456 [20]3 years ago
4 0

Alright, let's solve this problem! :)

<u>First we need to subtract 5k from both sides.</u>

3k−5k>12

<u>Then we simplify 3k - 5k to -2k.</u>

-2k > 12

<u>Now we divide both sides by -2.</u>

k < -12/2

<u>Now we simplify 12/2 to 6.  </u>

k < -6

✔️Done!

Have an amazing day!

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CaHeK987 [17]

Answer:

x =\$43.9

Step-by-step explanation:

The company will break even when it has made no profit, i.e. when y = 0:

y =0 = -x^2+48x-180.

Using the quadratic formula we get the two solutions:

x = \dfrac{-48\pm\sqrt{48^2-4(-1)(-180)}  }{-2}

which in decimal form are

x = \$4.1

x =\$43.9

We choose the price x =\$43.9, because it is the highest price for which no profit is made, and higher price means that you could sell least number of products to earn a certain amount of money.

Also. the graph of y(x) relates to the real life situation from the blue line shown in the graph, because what happens in real life is that as you increase the price, your profit decreases.

3 0
3 years ago
Given directed line segment PR below, find the coordinates of Q on PR
g100num [7]

Answer:

(\frac{4}{3},-3)

Step-by-step explanation:

The location of a point O(x, y) which divides line segment AB in the ratio a:b with point A at (x_1,y_1) and B(x_2,y_2) is given by the formula:

x=\frac{a}{a+b}(x_2-x_1)+x_1\\ \\y=\frac{a}{a+b}(y_2-y_1)+y_1

From the graph, the location of the point is P(2, 5), R(1, -7). Let point Q be (x, y) which divides PR in ratio 2:1. Coordinates of point Q is at:

x=\frac{a}{a+b}(x_2-x_1)+x_1=\frac{2}{2+1}(1-2)+2=\frac{2}{3}(-1)+2=\frac{4}{3}   \\ \\y=\frac{a}{a+b}(y_2-y_1)+y_1=\frac{2}{2+1}(-7-5)+5= \frac{2}{3}(-12)+5=-3

Point Q is at (\frac{4}{3},-3)

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3 years ago
Determin 16x - 4y + 2 = 0
Dafna11 [192]

Answer:

16inches is your answer

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3 years ago
3a+2x-3y=15<br>Solve for a.<br>Then Solve put the value for a =<br>a+7÷10​
erma4kov [3.2K]

\huge\bf Question:–

\sf \longmapsto \: 3a+2x−3y=15

\bf \huge \: To  \: Find:–

\boxed{\bf \: Value\: of  \: A}

\huge\bf Solution:–

\sf \longmapsto \: 3a+2x−3y=15

\boxed{ \bf \: Add -2x  \: to  \: both  \: sides}

\sf \longmapsto \: 3a+2x−3y+−2x=15+−2x

\sf \longmapsto \: 3a−3y=−2x+15

\boxed{ \bf \: Add  \: 3y  \: to \:  both  \: sides}

\sf \longmapsto \: 3a−3y+3y=−2x+15+3y

\sf \longmapsto \: 3a=−2x+3y+15

\boxed{\bf \:  \: Divide  \: both  \: sides \:  by \:  3}

\sf \longmapsto \:  \dfrac{3a}{3} =  \dfrac{−2x+3y+15}{3}

\boxed{\bf \:  Cross \: Multiply}

\boxed{\sf \longmapsto \: a =  \dfrac{ - 2}{3} x + y + 5}

______________________________________

\bf \: Put\:The\: Value

\sf \longmapsto \: \dfrac{−2x+3y+15}{3}  +7÷10

\boxed{\bf \: Distribute}

\sf \longmapsto \: \dfrac{ - 2}{3} x+y+5+ \dfrac{7}{10}

\boxed{\bf \: Combine \:  Like \:  terms}

\sf \longmapsto \: \bigg( \dfrac{ - 2}{3} x\bigg) + y +\bigg( 5 +  \dfrac{7}{10} \bigg)

\sf \longmapsto \:  \dfrac{ - 2}{3} x + y +  \dfrac{57}{10}

______________________________________

\boxed{\bf The Answer\: is:–}

\boxed{{\underline{\bf\dfrac{ - 2}{3} x + y +  \dfrac{57}{10}} }}

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