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MAXImum [283]
8 months ago
13

F(x) = 3x-1 , (f of g)(x) = x^2 , Find g(x)

Mathematics
1 answer:
fomenos8 months ago
6 0

Answer:

g(x)=\dfrac{x^2+1}{3}

Step-by-step explanation:

<u>Given functions</u>:

\begin{cases}f(x)=3x-1\\(f \circ g)(x)=x^2\end{cases}

Composite functions are when the <u>output</u> of one function is used as the <u>input</u> of another.

Therefore, the given <u>composite function</u> (f o g)(x) means to <u>substitute</u> function g(x) in place of the x in function f(x).

\begin{aligned}(f \circ g)(x) & = x^2\\f(g(x)) & = x^2\\\implies 3(g(x))-1 & = x^2\\3(g(x))-1+1 & = x^2+1\\3(g(x))&=x^2+1\\\dfrac{3(g(x))}{3}&=\dfrac{x^2+1}{3}\\\implies g(x)&=\dfrac{x^2+1}{3}\end{aligned}

Therefore:

\boxed{g(x)=\dfrac{x^2+1}{3}}

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Work out the surface area of this solid prism
IRISSAK [1]

Answer:

5220

Step-by-step explanation:

1- 20x36/2 is 360

(360 x 2 is 720)

2. 25x50 is 1250

3. 29x50 is 1450

4. 36x50 is 1800

720+1250+1450+1800 is 5220

3 0
2 years ago
Type the correct answer in each box. Use numerals instead of words.
lubasha [3.4K]

Answer:

System A has 4 real solutions.

System B has 0 real solutions.

System C has 2 real solutions

Step-by-step explanation:

System A:

x^2 + y^2 = 17   eq(1)

y = -1/2x            eq(2)

Putting value of y in eq(1)

x^2 +(-1/2x)^2 = 17

x^2 + 1/4x^2 = 17

5x^2/4 -17 =0

Using quadratic formula:

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

a = 5/4, b =0 and c = -17

x=\frac{-(0)\pm\sqrt{(0)^2-4(5/4)(-17)}}{2(5/4)}\\x=\frac{0\pm\sqrt{85}}{5/2}\\x=\frac{\pm\sqrt{85}}{5/2}\\x=\frac{\pm2\sqrt{85}}{5}

Finding value of y:

y = -1/2x

y=-1/2(\frac{\pm2\sqrt{85}}{5})

y=\frac{\pm\sqrt{85}}{5}

System A has 4 real solutions.

System B

y = x^2 -7x + 10    eq(1)

y = -6x + 5            eq(2)

Putting value of y of eq(2) in eq(1)

-6x + 5 = x^2 -7x + 10

=> x^2 -7x +6x +10 -5 = 0

x^2 -x +5 = 0

Using quadratic formula:

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

a= 1, b =-1 and c =5

x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(5)}}{2(1)}\\x=\frac{1\pm\sqrt{1-20}}{2}\\x=\frac{1\pm\sqrt{-19}}{2}\\x=\frac{1\pm\sqrt{19}i}{2}

Finding value of y:

y = -6x + 5

y = -6(\frac{1\pm\sqrt{19}i}{2})+5

Since terms containing i are complex numbers, so System B has no real solutions.

System B has 0 real solutions.

System C

y = -2x^2 + 9    eq(1)

8x - y = -17        eq(2)

Putting value of y in eq(2)

8x - (-2x^2+9) = -17

8x +2x^2-9 +17 = 0

2x^2 + 8x + 8 = 0

2x^2 +4x + 4x + 8 = 0

2x (x+2) +4 (x+2) = 0

(x+2)(2x+4) =0

x+2 = 0 and 2x + 4 =0

x = -2 and 2x = -4

x =-2 and x = -2

So, x = -2

Now, finding value of y:

8x - y = -17    

8(-2) - y = -17    

-16 -y = -17

-y = -17 + 16

-y = -1

y = 1

So, x= -2 and y = 1

System C has 2 real solutions

4 0
3 years ago
Read 2 more answers
If a-b/b = 3/7 What is b/a​
Basile [38]

Answer:

7/10

Step-by-step explanation:

Solve for  a  by simplifying both sides of the equation, then isolating the variable.

Glad I could help!

6 0
3 years ago
Convert 0.0000001 to a power of 10. 107 10−7 106 10−6
Westkost [7]
I think it's  10^-7 hope that helps
7 0
3 years ago
Read 2 more answers
Please answer ASAP. A baseball is hit upward from a platform that is m high at an initial speed of 29m/s. The approximate height
kotegsom [21]

Answer:

a) about 0.7 seconds to 5.1 seconds.

b) Listed below.

Step-by-step explanation:

h - 1 = -5x^2 + 29x

h = -5x^2 + 29x + 1

a) We will find the amount of time it takes to get to 18 meters.

18 = -5x^2 + 29x + 1

-5x^2 + 29x + 1 = 18

-5x^2 + 29x - 17 = 0

We will then use the quadratic formula to find the answer.

[please ignore the A-hat; that is a bug]

\frac{-29±\sqrt{29^2 - 4 * -5 * -17} }{2 * -5}

= \frac{-29±\sqrt{841 - 340} }{-10}

= \frac{-29±\sqrt{501} }{-10}

= \frac{-29 ± 22.38302929}{-10}

= \frac{-6.616970714}{-10} and \frac{-51.38302929}{-10}

= 0.6616970714 and 5.138302929

So, the time period for which the baseball is higher than 18 metres ranges from about 0.7 seconds to 5.1 seconds.

b) Restrictions on the domain and range of the function are that the domain and range can never be negative, since time cannot be negative, and height cannot be negative. The height cannot exceed the vertex of the parabola, since that is the highest the ball will ever go. It cannot exceed that height since gravity will cause the ball to fall down.

Hope this helps!

5 0
3 years ago
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