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Daniel [21]
2 years ago
11

F(x)=x-3 g(x)=x^2-x+1

Mathematics
1 answer:
Ainat [17]2 years ago
3 0

ansawer (-2x^2+8x-3)/(x-3)

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Given f(x) = 2x - 1, find the value of f(0).
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F(x) = 2x - 1

subtitute x = 0 to the equation:

f(0) = 2· 0 -1 = 0 - 1 = -1

f(0) = -1
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I will give out a Brainliest answer!!  Sylvia earns $7 per hour at her after school job. After working one week, she recieved a
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4. Simplify -6(3 – 5s + 12t – 2u)
Elan Coil [88]
-18 +30s -72t +12u is the answer
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2 years ago
The 10th term of an a.p is -27 and the 5th term is -12 .find the first term?​
horsena [70]

Step-by-step explanation:

a_{10} = - 27... (Given) \\\\\therefore a + 9d = - 27...... (1)\\\\a_{5} = - 12... (Given) \\\\\therefore a + 4d = - 12.......(2)\\\\

Subtracting equation (2) from equation (1)

a + 9d-( a + 4d )= - 27-(-12)\\\\\therefore a +9d - a-4d = - 27+12\\\\\therefore 5d = - 15\\\\\therefore d = \frac {- 15}{5}\\\\\huge \orange {\boxed {\therefore d = -3}} \\\\

Substituting d = - 3 in equation (1), we find:

a + 9\times (-3)= - 27\\\\\therefore a - 27 = - 27\\\\\therefore a  = 27- 27\\\\\huge \red {\boxed {\therefore a  = 0}}\\\\

Hence, first term is zero.

3 0
3 years ago
In Exercises 5-7, find all the exact t-values for which the given statement is true,
bearhunter [10]

Answer:  See Below

<u>Step-by-step explanation:</u>

NOTE: You need the Unit Circle to answer these (attached)

5) cos (t) = 1

Where on the Unit Circle does cos = 1?

Answer: at 0π (0°) and all rotations of 2π (360°)

In radians:     t = 0π + 2πn

In degrees:   t = 0° + 360n

******************************************************************************

6)\quad sin (t) = \dfrac{1}{2}

Where on the Unit Circle does   sin = \dfrac{1}{2}

<em>Hint: sin is only positive in Quadrants I and II</em>

\text{Answer: at}\  \dfrac{\pi}{6}\ (30^o)\ \text{and at}\ \dfrac{5\pi}{6}\ (150^o)\ \text{and all rotations of}\ 2\pi \ (360^o)

\text{In radians:}\ t = \dfrac{\pi}{6} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{6} + 2\pi n

In degrees:    t = 30° + 360n  and  150° + 360n

******************************************************************************

7)\quad tan (t) = -\sqrt3

Where on the Unit Circle does    \dfrac{sin}{cos} = \dfrac{-\sqrt3}{1}\ or\ \dfrac{\sqrt3}{-1}\quad \rightarrow \quad (1,-\sqrt3)\ or\ (-1, \sqrt3)

<em>Hint: sin and cos are only opposite signs in Quadrants II and IV</em>

\text{Answer: at}\  \dfrac{2\pi}{3}\ (120^o)\ \text{and at}\ \dfrac{5\pi}{3}\ (300^o)\ \text{and all rotations of}\ 2\pi \ (360^o)

\text{In radians:}\ t = \dfrac{2\pi}{3} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{3} + 2\pi n

In degrees:    t = 120° + 360n  and  300° + 360n

4 0
3 years ago
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