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Lina20 [59]
2 years ago
14

Solve the inequality. Graph the solution set. 26 + 6b 2(3b + 4)

Mathematics
2 answers:
Anna71 [15]2 years ago
7 0
<span>The inequality is : 26 + 6b= 2(3b + 4)
On opening the bracket we get,
or </span><span>26 + 6b= 6b + 8
Now, bring the variable terms to right hand side,
or 26 + 6b - 6b = 8
or 26 = 8, which is not possible because 26 can never be equal to 8.
Thus, this inequality has no solutions. </span>
brilliants [131]2 years ago
5 0

we have

26 + 6b\geq2(3b + 4)

Applying the distributive property on the right side

26 + 6b\geq6b+8

subtract 6b from both sides

26\geq 8 -------> is true

for all real numbers the inequality is true

therefore

the graph is a shaded area everywhere.

<u>the answer is</u>

the solution is all real numbers

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Answer:

That (blank) is 9.

Step-by-step explanation:

Please write 8x^2 - 48x = -104, with " ^ " indicating exponentiation.  Thanks.

This sounds like a perfect example of "completing the square."

First, factor 8x^2 - 48x = -104:  8(x^2 - 6x = -13).

If we focus SOLELY on x^2 - 6x = -13, it's just a little easier to "complete the square."  Take half of the coefficient of x (which here is -6), and square that result (which here is (-3)^2, or 9.  Now add 9 both sides of x^2 - 6x = -13:

x^2 - 6x + 9 = -13 + 9.  Note:  The "blank" in the problem statement is 9.  

We could go ahead and solve this, tho' we weren't asked to do so.  Rewrite x^2 - 6x + 9 as (x - 3)^2:

x^2 - 6x = -13 then becomes (x - 3)^2 = 4.  To solve for x, take the sqrt of both sides of this latest result, obtaining x - 3 = plus or minus 2.  Add 3 to both sides to isolate x:  x = 3 plus or minus 2.

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5 0
2 years ago
Read 2 more answers
What are the zeros of the quadratic function f(x) = 6x^2 + 12x – 7?
strojnjashka [21]

Answer:

x1=\frac{-2+\sqrt{26/3}}{2}

x2=\frac{-2-\sqrt{26/3} }{2}

Step-by-step explanation:

To find the zeros of the quadratic function f(x)=6x^2 + 12x – 7 we need to factorize the polynomial.

To do so, we need to use the quadratic formula, which states that the solution to any equation of the form ax^2 + bx + c = 0 is:

x=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

So, the first thing we're going to do is divide the whole function by 6:

6x^2 + 12x – 7 = 0 -> x^2 + 2x - 7/6

This step is optional, but it makes things quite easier.

Then we using the quadratic formula, where:

a=1, b= 2, c = -7/6.

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x=\frac{-2±\sqrt{2^{2}-4(1)(-7/6)}}{2}

x=\frac{-2±\sqrt{4 +14/3}}{2}

x=\frac{-2±\sqrt{26/3}}{2}

So the zeros are:

x1=\frac{-2+\sqrt{26/3}}{2}

x2=\frac{-2-\sqrt{26/3}}{2}

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2 years ago
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Answer:

Step-by-step explanation:

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s2008m [1.1K]

Answer:

B

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