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3 years ago

Solve this inequality: 3q + 11 + 8q > 99. A. q > 8 B. q > 1/8 C. q > 88 D. q > 11

1 answer:

3 0

Hey!

To solve this problem we must first group like terms.

Original Equation :
3q + 11 + 8q > 99

New Equation {Grouped Like Terms} :
3q + 8q + 11 > 99

Now, we'll add our similar elements.

Old Equation :
3q + 8q + 11 > 99

New Equation {Added Similar Elements} :
11q + 11 > 99

Next, we'll have to change the equation. To do this we'll add minus eleven to both sides of the equation. This will give us eleven q on its own.

Old Equation :
11q + 11 > 99

New Equation {Added Subtract 11 to Both Sides} :
11q + 11 - 11 > 99 - 11

Now we simplify.

Old Equation :
11q + 11 - 11 > 99 - 11

New Equation {Old Equation Simplified} :
11q > 88

Now, we'll have to divide both sides by eleven. We do this so as to get the variable ( q ) on its own.

Old Equation :
11q > 88

New Equation {Added Divide Both Sides by 11} :
 $\frac{11q}{11}$ > $\frac{88}{11}$

Once again, we now have to simplify.

Old Equation :
$\frac{11q}{11}$ > $\frac{88}{11}$

New Equation {Old Equation Simplified} :
q > 8

So, this means that the solution to the inequality given is q > 8.

Hope this helps!

- Lindsey Frazier ♥

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7th term of (c-d)^10

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$\bf (c-d)^{10}\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&+1&(c)^{10}(-d)^0\\
2&+10&(c)^9(-d)^1\\
3&+45&(c)^8(-d)^2\\
4&+120&(c)^7(-d)^3\\
5&+210&(c)^6(-d)^4\\
6&+252&(c)^5(-d)^5\\
7&+210&(c)^4(-d)^6
\end{array}$

as you notice, the first term from the binomial, starts off with a highest exponent of 10, and then gradually goes down by 1 on every element.

the second term of the binomial, starts off with an exponent of 0, and goes up by 1 on every element.

now, to get the coefficient
the first coefficient is 1 and the second is the same as the binomial's exponent... those two are simple.... now, to get any subsequent one, we simply "get the product of the current coefficient and the exponent of the first term, and divide it by the exponent of the second term on the next element".

for example... how do we get 10 for the second element, basically 1 * 10/1

that one is simple

now how do we get say, 45 for the 3rd element?
10 * 9 / 2

how do we get 252 for the 6th element?
210 * 6 / 5

and so on, and from there, you can expand each element.

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