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vlada-n [284]
3 years ago
6

Match the reasons with the statements. GIVEN: x2 + 6x + 2x + 12 = 0 TO PROVE: x = -6 or x = -2 1. x2 + 6x + 2x + 12 = 0 Subtract

ion property of equality 2. x2 + 8x + 12 = 0 Distributive Postulate 3. (x + 6)(x + 2) = 0 Combining like terms 4. x + 6 = 0 or x + 2 = 0 Zero product postulate 5. x = -6 or x = -2
Mathematics
2 answers:
mestny [16]3 years ago
6 0

Given: x^2 + 6x + 2x + 12 = 0.

1. Combining like terms (here terms 6x and 2x both contain x, then we can combine them):

x^2+(6x+2x)+12=0,\\\\x^2+8x+12=0.

2. Distributive postulate:

x^2+8x+12=(x+6)(x+2).

The equation is

(x+6)(x+2)=0.

3. Zero product postulate (zero product postulate state that if a product of two factors is equal to zero, then first factor is equal to zero or second factor is equal to zero):

x+6=0 \text{ or } x+2=0.

4. Subtraction property of equality:

a) subtract 6 from the first equation:

x+6=0\Rightarrow x+6-6=-6, \ x=-6.

b) subtract 2 from the second equation:

x+2=0\Rightarrow x+2-2=-2, \ x=-2.

xxTIMURxx [149]3 years ago
4 0

Answer:

1. x2 + 6x + 2x + 12 = 0 1. Given

2. x2 + 8x + 12 = 0        2. Combining like terms

3. (x + 6)(x + 2) = 0       3. Distributive Postulate

4. x + 6 = 0 or x + 2 = 0 4. Zero product postulate

5. x = -6 or x = -2         5 Subtraction property of equality


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vovikov84 [41]

There are 74.227 milliliters of acid in the solution.

Explanation:

  • The total amount of liquid is 373 milliliters. This is both the acid and water combined i.e amount of liquid + amount of acid = 373 milliliters
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  • 19.9% of 373 milliliters = 0.199 × 373 = 74.227 milliliters. So 74.227 milliliters of the entire solution is acid.

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3 years ago
Let N be the universal set and A, B, C, D be its subsets given by A={x:xisaevennaturalnumber},B={x:xNandxisamultipleof3} C={x:x
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Answer:

A^{'} = {1,3,5,7,9,....}

B^{'} = {1,2,4,5,7,8,10,11,13,14......}

C^{'} = {1,2,3,4,5}

D^{'} = {10,11,12,13,14,15,......}

Step-by-step explanation:

Solution:

It is given that, Universal set is the set of all Natural Numbers.

and

Set A, Set B, Set  C, Set D are subsets of Universal set.

A = {x:x is even natural numbers} = {2,4,6,8,10,.....}

B = {x:x ∈ N and x is a multiple of 3} = {3,6,9,12,15,....}

C = {x:x∈ N and x < 5} = {6,7,8,9,10,.....}

D = {x:x ∈ N and x > 10} = {1,2,3,4,5,6,7,8,9}

U = {1,2,3,4,5,6,7,8,9,10,11,........}

Complements:

A^{'} = U-A = {1,2,3,4,5,6,7,8,9,10,11,........} - {2,4,6,8,10,.....}

A^{'} = {1,3,5,7,9,....}

B^{'} = U-B =  {1,2,3,4,5,6,7,8,9,10,11,........} - {3,6,9,12,15,....}

B^{'} = {1,2,4,5,7,8,10,11,13,14......}

C^{'} = U-C = {1,2,3,4,5,6,7,8,9,10,11,........} - {6,7,8,9,10,.....}

C^{'} = {1,2,3,4,5}

D^{'} = {1,2,3,4,5,6,7,8,9,10,11,........} - {1,2,3,4,5,6,7,8,9}

D^{'} = {10,11,12,13,14,15,......}

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