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2 years ago

Which point is a solution to the inequality shown in this graph

Mathematics

2 answers:

maw [93]2 years ago

5 0

A solution would be (0,5)

it cannot be (3,3) or (-3,-1) because u have a dashed line...and it cannot be (0,0) because that is not in the solution area either

yuradex [85]2 years ago

3 0

Step 1

Find the equation of the line of the inequality

Let

$A(-3,-1)\ B(3,3)$

Find the slope of the line

The slope is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{3+1}{3+3}$

$m=\frac{4}{6}$

$m=\frac{2}{3}$

Find the equation of the line into point-slope form

$y-y1=m(x-x1)$

we have

$m=\frac{2}{3}$

$(x1,y1)=B(3,3)$

substitute in the equation

$y-3=\frac{2}{3}(x-3)$

$y=\frac{2}{3}x-2+3$

$y=\frac{2}{3}x+1$

Find the equation of the inequality

The solution is the shaded area above the dotted line

so the inequality is

$y>\frac{2}{3}x+1$

If a point is the solution of the inequality, then it must satisfy the inequality. Let's check each of the points

Step 2

case A) $(0,0)$

Substitute the values of x and y in the inequality

$x=0\ y=0$

$0>\frac{2}{3}*0+1$

$0>1$ -----> is not true

therefore

the point $(0,0)$ is not solution of the inequality

Step 3

case B) $(3,3)$

Substitute the values of x and y in the inequality

$x=3\ y=3$

$3>\frac{2}{3}*3+1$

$3>3$ -----> is not true

therefore

the point $(3,3)$ is not solution of the inequality

Step 4

case C) $(-3,-1)$

Substitute the values of x and y in the inequality

$x=-3\ y=-1$

$-1>\frac{2}{3}*-3+1$

$-1>-1$ -----> is not true

therefore

the point $(-3,-1)$ is not solution of the inequality

Step 5

case D) $(0,5)$

Substitute the values of x and y in the inequality

$x=0\ y=5$

$5>\frac{2}{3}*0+1$

$5>1$ -----> is true

therefore

the point $(0,5)$ is a solution of the inequality

therefore

the answer is

$(0,5)$

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2 years ago

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Answer:

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2 years ago

A store sells 8 colors of balloons with at least 28 of each color. How many different combinations of 28 balloons can be chosen?

Len [333]

Answer:

(a) $Selection = 6724520$

(b) $At\ most\ 12 = 6553976$

(c) $At\ most\ 8 = 6066720$

(d) $At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

Step-by-step explanation:

Given

$Colors = 8$

$Balloons = 28$ --- at least

Solving (a): 28 combinations

From the question, we understand that; a combination of 28 is to be selected. Because the order is not important, we make use of combination.

Also, because repetition is allowed; different balloons of the same kind can be selected over and over again.

So:

$n => 28 + 8-1$ $= 35$

$r = 28$

$Selection = ^{35}^C_{28$

$Selection = \frac{35!}{(35 - 28)!28!}$

$Selection = \frac{35!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29*28!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29}{7!}$

$Selection = \frac{35*34*33*32*31*30*29}{7*6*5*4*3*2*1}$

$Selection = \frac{33891580800}{5040}$

$Selection = 6724520$

Solving (b): At most 12 red balloons

First, we calculate the ways of selecting at least 13 balloons

Out of the 28 balloons, there are 15 balloons remaining (i.e. 28 - 13)

So:

$n => 15 + 8 -1 = 22$

$r = 15$

Selection of at least 13 =

$At\ least\ 13 = ^{22}C_{15}$

$At\ least\ 13 = \frac{22!}{(22-15)!15!}$

$At\ least\ 13 = \frac{22!}{7!15!}$

$At\ least\ 13 = 170544$

Ways of selecting at most 12 =

$At\ most\ 12 = Total - At\ least\ 13$ --- Complement rule

$At\ most\ 12 = 6724520- 170544$

$At\ most\ 12 = 6553976$

Solving (c): At most 8 blue balloons

First, we calculate the ways of selecting at least 9 balloons

Out of the 28 balloons, there are 19 balloons remaining (i.e. 28 - 9)

So:

$n => 19+ 8 -1 = 26$

$r = 19$

Selection of at least 9 =

$At\ least\ 9 = ^{26}C_{19}$

$At\ least\ 9 = \frac{26!}{(26-19)!19!}$

$At\ least\ 9 = \frac{26!}{7!19!}$

$At\ least\ 9 = 657800$

Ways of selecting at most 8 =

$At\ most\ 8 = Total - At\ least\ 9$ --- Complement rule

$At\ most\ 8 = 6724520- 657800$

$At\ most\ 8 = 6066720$

Solving (d): 12 red and 8 blue balloons

First, we calculate the ways for selecting 13 red balloons and 9 blue balloons

Out of the 28 balloons, there are 6 balloons remaining (i.e. 28 - 13 - 9)

So:

$n =6+6-1 = 11$

$r = 6$

Selection =

$^{11}C_6 = \frac{11!}{(11-6)!6!}$

$^{11}C_6 = \frac{11!}{5!6!}$

$^{11}C_6 = 462$

Using inclusion/exclusion rule of two sets:

$Selection = At\ most\ 12 + At\ most\ 8 - (12\ red\ and\ 8\ blue)$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 170544+ 657800- 462$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = Total - Only\ 12\ red\ and\ only\ 8\ blue$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 6724520 - 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

3 0

2 years ago