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bogdanovich [222]
3 years ago
15

F(x) =x^2+6x and g(x)=7-x find (f-g)(x) and (f-g)(9)

Mathematics
1 answer:
liq [111]3 years ago
8 0
(f-g)(x) = x²+6x - (7-x)
(f-g)(x) = x² + 6x - 7 + x
(f-g)(x) = x² + 7x - 7

(f-g)(9) = 9² + 7*9 - 7
(f-g)(9) = 81 + 63 - 7
(f-g)(9) = 137
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Using the quadratic formula, what is the value of x for x^2-4+5=0​
PSYCHO15rus [73]

value of x is: x= i and x= -i

Step-by-step explanation:

We need to find the value of x from x^2-4+5=0​ using quadratic formula.

The term is: x^2+1=0​

The quadratic formula is:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\

Where a = 1, b=0,c=1

Putting values:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{0\pm\sqrt{(0)^2-4(1)(1)}}{2(1)}\\x=\frac{\pm\sqrt{-4}}{2}\\x=\frac{\pm2i}{2}\\x=\pm i

So, value of x is: x= i and x= -i

Keywords: Quadratic formula

Learn more about quadratic formula at:

  • brainly.com/question/7361044
  • brainly.com/question/10666510
  • brainly.com/question/1332667

#learnwithBrainly

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4 years ago
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Step-by-step explanation:

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3 years ago
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daser333 [38]
To complete the identity, we need these fundamental identities:

1)\displaystyle{sec(x)=\frac{1}{cos(x)}

2) cos(x-y)=cos(x)cos(y)+sin(x)sin(y)

\displaystyle{csc(x)= \frac{1}{sin(x)}


Thus, by identity 1 we have:

\displaystyle{ sec( \frac{ \pi }{2}-\theta )= \frac{1}{cos(\frac{ \pi }{2}-\theta)}

by identity :

\displaystyle{cos(\frac{ \pi }{2}-\theta)=cos(\frac{ \pi }{2})cos(\theta)+sin(\frac{ \pi }{2})sin(\theta)

recall the values :

\displaystyle{ sin(\frac{ \pi }{2})^R=sin(90^o)=1\\\\

\displaystyle{ cos(\frac{ \pi }{2})^R=cos(90^o)=0, 


so: 

cos(\frac{ \pi }{2})cos(\theta)+sin(\frac{ \pi }{2})sin(\theta)=0+sin(\theta)=sin(\theta)


Putting all these together, we have:


\displaystyle{ sec( \frac{ \pi }{2}-\theta )= \frac{1}{cos(\frac{ \pi }{2}-\theta)}= \frac{1}{cos(\frac{ \pi }{2})cos(\theta)+sin(\frac{ \pi }{2})sin(\theta)}= \frac{1}{sin(\theta)}}

which is equal to csc(\theta), by identity 3


Answer: D
7 0
4 years ago
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