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Alecsey [184]
3 years ago
15

Let f(x) = 12 over the quantity of 4 x + 2. Find f(−1). (1 point)

Mathematics
1 answer:
mylen [45]3 years ago
6 0

Answer:

-6

Step-by-step explanation:

Hope I understand your "f(x) = 12 over the quantity of 4 x + 2" correctly

f(x) = 12 / 4x +2

Now so they want a f(-1), as u can see they substituted the x for - 1, therefor do that the same to the right side

f(-1) = 12/ 4(-1) + 2

f(-1) = 12 / - 4+2

= 12/ - 2

= - 6

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Help please thank you
butalik [34]
A - 2x^2 + 2x - 2

To find this, set up the equation:

(-x^2 + 6x - 1) + ( 3x^2 - 4x - 1)

With this, you need to combine like terms while taking into consideration the addition sign.

-x^2 + 3x^2 = 2x^2
6x + - 4x = 2x
- 1 + - 1 = - 2

Hope this helps!




7 0
3 years ago
Math. My brain cells have officially ADIOSED out of my brain.
natima [27]

Answer:

Step-by-step explanation:  40

4 0
3 years ago
173 out of 400 pages what is the percent he read
Lubov Fominskaja [6]
173/400 pages of the percentage
6 0
3 years ago
If a = √3-√11 and b = 1 /a, then find a² - b²​
Serga [27]

If b=\frac1a, then by rationalizing the denominator we can rewrite

b = \dfrac1{\sqrt3-\sqrt{11}} \times \dfrac{\sqrt3+\sqrt{11}}{\sqrt3+\sqrt{11}} = \dfrac{\sqrt3+\sqrt{11}}{\left(\sqrt3\right)^2-\left(\sqrt{11}\right)^2} = -\dfrac{\sqrt3+\sqrt{11}}8

Now,

a^2 - b^2 = (a-b) (a+b)

and

a - b = \sqrt3 - \sqrt{11} + \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{9\sqrt3 - 7\sqrt{11}}8

a + b = \sqrt3 - \sqrt{11} - \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{7\sqrt3 - 9\sqrt{11}}8

\implies a^2 - b^2 = \dfrac{\left(9\sqrt3 - 7\sqrt{11}\right) \left(7\sqrt3 - 9\sqrt{11}\right)}{64} = \boxed{\dfrac{441 - 65\sqrt{33}}{32}}

5 0
2 years ago
5 1/3 + ( 2 1/3 + 1 1/3)=
Fiesta28 [93]
Okay so you want to do the parentheses first okay? So 2.3333...+1.3333333 is?3.666 or 3 2/3. So you got that right? So then 5 1/3+ 3 2/3. You get an easy 9. Because 1/3+ 2/3 adds up to 3/3, and is normally written as 1. So add a one to 5+3=8+1.
6 0
4 years ago
Read 2 more answers
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