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3 years ago

Use the functions a(x) = 4x + 9 and b(x) = 3x − 5 to complete the function operations listed below.

1 answer:

4 0

A. (a + b)(x) =a(x)+b(x) = 4x + 9+ 3x − 5 = 7x - 4
B. (a ⋅ b)(x)= a(x)*b(x) =(4x + 9)( 3x − 5 )= 12x²+27x-20x-45 = 12x² +7x -45
C. a[b(x)] = 4(3x − 5) + 9 = 12x-20+9=12x-11

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-5x -2y= -9 8x + 8y= 24

max2010maxim [7]

Answer:

x=5

y=-17

Step-by-step explanation:

basically what I did was:

1) I multiplied the first equation by 4 , solved for x

2) I used the value of x that i found and substituted in one of the equations to find y

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3 years ago

Solve 2x2 - 6x = -8.

erma4kov [3.2K]

Answer:

x=2

Step-by-step explanation:

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3 years ago

I only need help with 12 and please tell me how you figured it out

Mars2501 [29]

18/3=6
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4 years ago

Find the exact value, without a

alexira [117]

Answer:

$-2 -\sqrt{3}$

Step-by-step explanation:

First consider numerator

$sin \frac{\frac{7\pi}{6}}{2} = sin \frac{7\pi}{12}= sin (\frac{\pi}{4} + \frac{\pi}{3})$

Using the formula : sin (A + B) = sin A cos B + cos A sin B

$sin \frac{\pi}{4} = \frac{\sqrt{2} }{2}, \ cos \frac{\pi}{4} = \frac{\sqrt{2} }{2}\\\\sin \frac{\pi}{3} = \frac{\sqrt{2} }{2}, \ cos \frac{\pi}{3} = \frac{1}{2}$

$sin(\frac{\pi}{4} + \frac{\pi}{3}) = sin \frac{\pi}{4} \cdot cos \frac{\pi}{3} + cos \frac{\pi}{4} \cdot sin \frac{\pi}{3}$

$=\frac{\sqrt{2} }{2} \cdot \frac{1}{2} + \frac{\sqrt{2} }{2} \cdot \frac{\sqrt{3} }{2} \\\\= \frac{\sqrt{2} }{4} + \frac{\sqrt{6} }{4}\\\\=\frac{\sqrt{2} +\sqrt{6} }{4}$

Second consider denominator

$cos \frac{\frac{7\pi}{6}}{2} = cos \frac{7\pi}{12}= cos (\frac{\pi}{4} + \frac{\pi}{3})$

Using the formula : cos (A + B) = cos A cos B - sin A sin B

$sin \frac{\pi}{4} = \frac{\sqrt{2} }{2}, \ cos \frac{\pi}{4} = \frac{\sqrt{2} }{2}\\\\sin \frac{\pi}{3} = \frac{\sqrt{2} }{2}, \ cos \frac{\pi}{3} = \frac{1}{2}$

$cos(\frac{\pi}{4} + \frac{\pi}{3}) = cos \frac{\pi}{4} \cdot cos \frac{\pi}{3} -sin \frac{\pi}{4} \cdot sin \frac{\pi}{3}$

$=\frac{\sqrt{2} }{2} \cdot \frac{1}{2} - \frac{\sqrt{2} }{2} \cdot \frac{\sqrt{3} }{2}\\\\=\frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4}\\\\= \frac{\sqrt{2} -\sqrt{6} }{4}$

Therefore,

$tan \frac{7\pi}{12} = \frac{sin \frac{7\pi}{12}}{cos\frac{7\pi}{12}}$

$= \frac{\frac{\sqrt{2} +\sqrt{6} }{4}} {\frac{\sqrt{2} -\sqrt{6} }{4} }\\\\=\frac{\sqrt{2} +\sqrt{6} }{4} \times \frac{4 }{\sqrt{2} -\sqrt{6}}\\\\=\frac{\sqrt{2} +\sqrt{6} }{\sqrt{2} -\sqrt{6}}$

Either we can stop here or Rationalize the denominator:

$\frac{\sqrt{2} +\sqrt{6} }{\sqrt{2} -\sqrt{6}} \times \frac{\sqrt{2} +\sqrt{6} }{\sqrt{2} +\sqrt{6}} = \frac{(\sqrt{2} +\sqrt{6})^{2} }{(\sqrt{2})^2 -(\sqrt{6})^2} = \frac{2 + 6 +2\sqrt{12} }{2-6} = \frac{8+2\sqrt{12} }{-4} = \frac{8+ 4\sqrt{3} }{-4} = -2-\sqrt{3}$

3 0

3 years ago

What is the surface area of the rectangular pyramid below? 13,13,13

NARA [144]

Answer:

1,014

Step-by-step explanation:

Assuming those 13’s represent the Length and Width of the sides of the prism, the equation would be

13x13x6 (6 sides)

=1014

5 0

4 years ago