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KATRIN_1 [288]
4 years ago
12

I have a few questions I need help on.

Mathematics
2 answers:
nika2105 [10]4 years ago
4 0
X= 4, x=-2
121/4
x=8, x=-12
There is a solution but it's not one of the options unless I'm wrong....
8 in 


Bumek [7]4 years ago
4 0
1.) 3x² - 6x - 24 = 0
x² - 2x - 8 = 0
x² + 2x - 4x - 8 = 0
x(x + 2) -4(x + 2) = 0
(x - 4) (x + 2) = 0
x = 4 OR -2

In short, Your Answer would be: Option C

2.) Expression: x² + 11x + n 
For perfect square trinomial, c = (b/2)²
c = (11/2)² = (5.5)² =30.25

In short, Your Answer would be: Option C

3.) x² + 4x - 96 = 0
x² - 8x + 12x - 96 = 0
x(x - 8) 12(x - 8) = 0
(x+12) (x-8) = 0
x = -12 OR 8

In short, Your Answer would be: Option B

4.) D.) There is no solution.

5.) We know, V = l * w * h
w * l = 2720/17 = 160

Let, the width = x
Length = 2x + 4
Then, (x)(2x + 4) = 160
2x² + 4x - 160 = 0
x² + 2x - 80 = 0
x² -8x + 10x - 80 = 0
x(x - 8) 10(x - 8) = 0
(x - 8) (x + 10) = 0
x = 8 OR -10
Length can't be in negative so, -10 is rejected.

Width = 8
Length = 2(8) + 4 = 16+4 = 20

In short, Your Answer would be: Option B

Hope this helps!
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5 0
4 years ago
Read 2 more answers
2. Lab groups of three are to be randomly formed (without replacement) from a class that contains five engineers and four non-en
Anna11 [10]

Answer:

The number of different lab groups possible is 84.

Step-by-step explanation:

<u>Given</u>:

A class consists of 5 engineers and 4 non-engineers.

A lab groups of 3 are to be formed of these 9 students.

The problem can be solved using combinations.

Combinations is the number of ways to select <em>k</em> items from a group of <em>n</em> items without replacement. The order of the arrangement does not matter in combinations.

The combination of <em>k</em> items from <em>n</em> items is: {n\choose k}=\frac{n!}{k!(n-k)!}

Compute the number of different lab groups possible as follows:

The number of ways of selecting 3 students from 9 is = {n\choose k}={9\choose 3}

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Thus, the number of different lab groups possible is 84.

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