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sleet_krkn [62]
3 years ago
11

Write the quadratic equation in general form and then choose the value of "b."

Mathematics
1 answer:
eduard3 years ago
3 0
The answer is b. 11. 

You can find this by multiplying the two parenthesis together. In order to do this, you can use a multiplying process known as FOIL (First, Outer, Inner, Last), where you multiply parts of the parenthesis in this order to make sure you do so properly. 

First Numbers:
2x * x = 2x^2

Outer Numbers: 
2x * 6 = 12x

Inner Numbers: 
-1 * x = -1x

Last Numbers: 
-1*6 = -6

Now place them all in that order and simplify. 
2x^2 + 12x - x - 6 ---> simplify x terms
2x^2 + 11x - 6

Since the middle term (the one with the x value) has a coefficient of 11, that gives us a b value of 11. 
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\left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
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-------------------------------\\\\

\bf \cfrac{\left( 81x^3y^{-\frac{1}{3}} \right)^{\frac{1}{4}}}{(x^2y)^{\frac{1}{4}}}\implies \cfrac{81^{\frac{1}{4}}x^{3\cdot {\frac{1}{4}}}y^{-\frac{1}{3}\cdot {\frac{1}{4}}}}{x^{2\cdot {\frac{1}{4}}}y^{\frac{1}{4}}}\implies
\cfrac{81^{\frac{1}{4}}x^{\frac{3}{4}}y^{-\frac{1}{12}}}{x^{\frac{2}{4}} y^{\frac{1}{4}}}

\bf \cfrac{81^{\frac{1}{4}}x^{\frac{3}{4}}x^{-\frac{2}{4}}}{ y^{\frac{1}{4}}y^{\frac{1}{12}}}\implies \cfrac{(3^4)^{\frac{1}{4}}x^{\frac{3}{4}-\frac{2}{4}}}{y^{\frac{1}{4}+\frac{1}{12}}}\implies \cfrac{3^{4\cdot \frac{1}{4}}x^{\frac{1}{4}}}{y^{\frac{4}{12}}}\implies \cfrac{3^{\frac{4}{4}}x^{\frac{1}{4}}}{y^{\frac{1}{3}}}
\\\\\\
\cfrac{3\sqrt[4]{x}}{\sqrt[3]{y}}
7 0
3 years ago
A box contains 5 red balls, 7 green balls, and 6 yellow balls. In how many ways can 6 balls be chosen if there should be 2 balls
Nostrana [21]

Answer:

12 balls

Step-by-step explanation:

take 3 red balls ,5 green balls and 4 yellow balls that 12 balls = 6 balls in bag

5 0
3 years ago
Which one is it?<br> A, B, C, or D?
grigory [225]

Answer: I'd go with C

Step-by-step explanation:

3 0
3 years ago
Which algebraic expression is equivalent to the expression below?
prohojiy [21]
You can again ignore the parenthesis because you are not distributing anything. 
Your equation will look like this

3x + 11 + 6x

You can move each of these numbers around any way you like. You can combine the 3x and the 6x if you want, but they did not do that. You cannot take the x away and put it in front of the 11 though.

B. is your answer. All they did was move the 6x inside the parenthesis and the 11 out of the parenthesis. 

Always remember, when you are adding things together, the parenthesis don't matter! 
3 0
3 years ago
Read 2 more answers
A. Let a, b, c be positive integers and suppose that
kodGreya [7K]

Answer:

Step-by-step explanation:

a|c means that c=a*k k is some positive integer. We know that b|c so b| ak and (a,b)=1, so it must be b|k, i.e k=b*r, r is some positive integer number. Now we have that c=abr, so ab| c.

B) if x and x’ are both solution then we have that

mi | x-x’ for every i.

By a) we have that m1m2...mk| x-x’, so x and x’ are equal by mod od m1m2...mk.

5 0
3 years ago
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