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3 years ago

Is (-2 , 3) a solution of the graphed inequality?

Mathematics

1 answer:

Sliva [168]3 years ago

7 0

Answer:

If it is outside of the shaded area, no

If it is inside the shaded area, yes

good luck, hope this helps :)

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Almost sooo close guys just 5 more left sorry if it’s taking long

Andrei [34K]

Answer:

As a decimal its 6.34, If you round up then its 6.35

As a fraction its 400/63

As a mixed number its 6 22/63

5 0

3 years ago

One urn contains one blue ball (labeled B1) and three red balls (labeled R1, R2, and R3). A second urn contains two red balls (R

marusya05 [52]

Answer:

(a) See attachment for tree diagram

(b) 24 possible outcomes

Step-by-step explanation:

Given

$Urn\ 1 = \{B_1, R_1, R_2, R_3\}$

$Urn\ 2 = \{R_4, R_5, B_2, B_3\}$

Solving (a): A possibility tree

If urn 1 is selected, the following selection exists:

$B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]$

If urn 2 is selected, the following selection exists:

$B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]$

See attachment for possibility tree

Solving (b): The total number of outcome

For urn 1

There are 4 balls in urn 1

$n = \{B_1,R_1,R_2,R_3\}$

Each of the balls has 3 subsets. i.e.

$B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]$

So, the selection is:

$Urn\ 1 = 4 * 3$

$Urn\ 1 = 12$

For urn 2

There are 4 balls in urn 2

$n = \{B_2,B_3,R_4,R_5\}$

Each of the balls has 3 subsets. i.e.

$B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]$

So, the selection is:

$Urn\ 2 = 4 * 3$

$Urn\ 2 = 12$

Total number of outcomes is:

$Total = Urn\ 1 + Urn\ 2$

$Total = 12 + 12$

$Total = 24$

5 0

2 years ago

Herr is the question and pic for the last question its 2

nevsk [136]

To find the area of the triangle you would have to plug in the numbers for this equation: A= Hxb/2
So in your case 14x10/2
140/2= 70 and that would be your answer. Answer:G

3 0

3 years ago

If LM = MK, LK =7x-10, KN = x+3, MN = 9x-11, and KJ =28, find LJ

marusya05 [52]

Answer:

Step-by-step explanation:

Suppose J, K, L, M, N are points on the same line.

MK = MN + (-KN) = MN - KN = 9x - 11 - x - 3 = 8x - 14

Since LK = MK and LK = 7x - 10, then

7x - 10 = 8x - 14

8x - 7x = -10 + 14

x = 4

LJ = MK + KJ

MK = LK = 7x - 10 = 7(4) - 10 = 28 - 10 = 18

LJ = 18 + 28 = 46

Read more on Brainly.com - brainly.com/question/5067118#readmore

8 0

3 years ago

Read 2 more answers

To eliminate the y terms and solve for x in the fewest steps, by which constants should the equation need multiplied by before a

ivanzaharov [21]

A. Multiplied by 2 for the first then for the second multiply by three.

Explanation: the equations already have opposite signs for y so reversing the signs would be counterproductive. (Eliminates B and D) and the easiest way is to get the numbers before the Y variable to be the same. The top by 2 and bottom by 3 would do that, leaving only A.

3 0

2 years ago