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11111nata11111 [884]
3 years ago
9

F(x) = 3x + 9, g(x) = 3x2 Find (fg)(x).

Mathematics
1 answer:
Brrunno [24]3 years ago
5 0
See picture for full reply.

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PLEASE HELP DUE IN 40 MINS!! Using the Law of Sines, find the exact answer for x.
Savatey [412]

Answer:

Step-by-step explanation:

Sin(30) / X  = Sin(45) / 12

use algebra to isolate X

1 / X = [ Sin(45) / 12  ]   /  Sin(30)

flip both sides

X =  Sin(30) / [Sin(45) /12 ]

invert the denominator on the right side

X = Sin(30) *  12/ Sin(45)

Do you know what Sin(30) is off the top of your head?

and also Sin(45) ?

these are worth remembering... and you can b/c they are just

sin(0) = 0/2

sin(30) = 1 / 2

sin(45) = \sqrt{2}/2

sin(60)= \sqrt{3} / 2

sin(90) = \sqrt{4} / 2    (aka 1)

note the numerators just counts up  0, 1, 2, 3, 4  :)  

Cos works the same way but counts from 90° back to 0 but exactly like sin other wise, hence why Cos and Sin both = \sqrt{2}/2 at 45 °

anyway

X =  1 / 2 * 12/ \sqrt{2}/2

X = 3\sqrt{2}

A=3

B= 2

:)  nice , huh

8 0
3 years ago
-6, -5, -3, -1,... what comes next?
lys-0071 [83]

Answer:

1

Step-by-step explanation:

6 0
3 years ago
A game designer must decide how to color five buildings that are in a row. Using only the colors yellow, green, red, and blue, e
earnstyle [38]

Answer:

The total of ways the buildings can be painted are  = 4 \times 3  \times 2  \times 2  \times 1 = 48 ways.

Step-by-step explanation:

i) color five buildings that are in a row.

ii) any two neighboring buildings must be different colors

iii) the first third and fifth buildings must be different colors.

iv) the number of ways to paint the fist building are 4.

v) the number of ways to paint the second building are 3.

vi) the number of ways to paint the third building are 2.

vii) the number of ways to paint the fourth building are 2.

viii) the number of ways to paint the last building are 1.

ix) therefore the total of ways the buildings can be painted are

   = 4 \times 3  \times 2  \times 2  \times 1 = 48 ways.

4 0
3 years ago
Find two consecutive integers such that the larger is nine more than twice the smaller
ivann1987 [24]
A=b+1
a=2b+9

b+1=2b+9
-b from both sides
1=b+9
-9 from both sides
-8=b

b+1
-8+1
-7

You can check your answer...
2b+9
2(-8)+9
-16+9
-7

The smaller is -8 and the larger is -7
6 0
4 years ago
What is the completely factored form of p4 – 16?
Svetlanka [38]

Answer:

(p+2)(p-2)(p²+4)

Step-by-step explanation:

The binomial p⁴-16 is a difference of squares.  This is because p⁴ is a perfect square, as is 16.  This means we can write it as the sum of the square roots multiplied by the difference of the square roots:

(p²+4)(p²-4)

The first binomial, p²+4, cannot be factored any further.

The second binomial, p²-4, is another difference of squares, since p² is a perfect square, as is 4:

(p+2)(p-2)

This gives us

(p²+4)(p+2)(p-2)

8 0
3 years ago
Read 2 more answers
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