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slava [35]
3 years ago
9

Given: f(x) = 3 - X; g(x) = -2x Find g[f(x)]

Mathematics
1 answer:
OLEGan [10]3 years ago
5 0

Answer:

g[f(x)] = -2(3 - x) = -6 + 2x

Step-by-step explanation:

When we substitute the function f(x) for x in g(x) = -2x, we get the composition g[f(x)]:

g[f(x)] = -2(3 - x) = -6 + 2x

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If you could only plot 5 points to graph y=sinx, which points would you choose? How would you connect these points? Please expla
fenix001 [56]

Answer:

For both graph I would choose the following points: 0°, 90°, 180°, 270° and 360°. I would connect the points (in both cases) with an smooth arc.

Step-by-step explanation:

x          sin(x)     cos(x)

0°        0            1

90°      1             0

180°     0           -1

270°    -1            0

360°    0            1

8 0
3 years ago
Will mark BRAINLIEST if you just answer this one question and show your work, thank you! please help asap​
Natalka [10]

On a die there are 3 even numbers. We want to know the probability that rolling a die we get 3 or an even number. So there are 4 positive outcomes out of 6.

4/6=2/3

3 0
3 years ago
Can to answer all 4 please?
Tasya [4]
1) A - yes, b - no, c -  yes
2) false, true, false
3) 85p - 34%
      xp - 66%
x = 165 - she has to read 165 more pages.
4) Sylvia earns 84/12 = 7$/hr
maximum she can earn is 7*15 = 105 $
she will earn 105$ if she works for 15 hours
5 0
2 years ago
The straight line 2x + 3y = 7 meets the curve
NISA [10]

Step-by-step explanation:

xy + 10 = 0. => y = -10/x.

2x + 3y = 7. => 3y = -2x + 7, y = -2/3 x + 7/3.

When -10/x = -2/3 x + 7/3,

-10 = -2/3 x² + 7/3 x, 2/3 x² - 7/3 x - 10 = 0.

=> 2x² - 7x - 30 = 0

=> (2x + 5)(x - 6) = 0

=> x = -2.5 or x = 6.

dy/dx = d/dx [-10/x] = 10/x².

When x = -2.5, dy/dx = 10 / (-2.5)² = 1.6.

When x = 6, dy/dx = 10 / (6)² = 5/18.

Hence the gradients at the points are 1.6 and 5/18.

5 0
3 years ago
HELP ME QUICK PLEASEEEE solve for x: 2/x-2+7/x^2-4=5/x
Anna35 [415]

 

\displaystyle\\\\\frac{2}{x-2}+\frac{7}{x^2-4}=\frac{5}{x}\\\\\frac{^{x+2)}2~~~~~}{x-2}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2(x+2)}{(x-2)(x+2)}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2x+4+7}{x^2-4}=\frac{5}{x}\\\\\frac{2x+11}{x^2-4}=\frac{5}{x}\\\\5(x^2-4)=x(2x+11)\\\\5x^2-20=2x^2+11x\\\\5x^2-2x^2 -11x-20=0


\displaystyle\\3x^2-11x-20=0\\\\x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{11\pm\sqrt{121+4\cdot3\cdot20}}{2\cdot3}=\\\\=\frac{11\pm\sqrt{121+240}}{6}=\frac{11\pm\sqrt{361 }}{6}=\frac{11\pm19}{6}\\\\x_1=\frac{11-19}{6}=\frac{-8}{6}\\\\\boxed{\bf x_1=-\frac{4}{3}}\\\\x_2=\frac{11+19}{6}=\frac{30}{6}\\\\\boxed{\bf x_2=5}




3 0
3 years ago
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