Someone answer the empty ones

Mathematics

1 answer:

Mkey [24]3 years ago

7 0

The addition would be 1.38

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Which function is equivalent to f(x)=1/4(x-2)^2+3?

QveST [7]

Answer: f(6) = (6)2 + 1 = 36 + 1 = 37

If we wanted to find and expression for f(g(x)) we could plug in g(x), which is "x - 4" into the function f(x) wherever we see an "x". So:

f(x) = x2 + 1

f(g(x)) = (x - 4)2 + 1 because g(x) is "x - 4"

Step-by-step explanation:

8 0

3 years ago

In the expansion of (1/ax +2ax^2)^5 the coefficient of x is five. Find the value of the constant a.

DedPeter [7]

Answer:

80x⁴

Step-by-step explanation:

$(\frac{1}{ax} + 2ax^2)^5 = 5C_0(\frac{1}{ax})^5(2ax^2)^0 + 5C_1(\frac{1}{ax})^4(2ax^2)^1 + 5C_2(\frac{1}{ax})^3 (2ax^2)^2$

$+ 5C_3 (\frac{1}{ax})^2(2ax^2)^3 + 5C_4(\frac{1}{ax})^1(2ax^2)^4 + 5C_5(\frac{1}{ax})^0(2ax^2)^5$

$5C_0(\frac{1}{ax})^5(2ax^2)^0 =1 \times (\frac{1}{ax})^5 \times 1 = \frac{1}{a^5x^5}\\\\5C_1(\frac{1}{ax})^4(2ax^2)^1 = 5 \times (\frac{1}{ax})^4 \times (2ax^2)^1 = 10 ax^2 \times \frac{1}{a^4x^4} = \frac{10}{a^3x^2}\\\\5C_2 (\frac{1}{ax})^3 (2ax^2)^2= 10 \times (\frac{1}{ax})^3 \times (2ax^2)^2 = 10 \times \frac{1}{a^3x^3} \times 4a^2x^4 = \frac{40x}{a}\\\\5C_3 (\frac{1}{ax})^2 (2ax^2)^3 = 10 \times (\frac{1}{ax})^2 \times (2ax^2)^3 = 10 \times \frac{1}{a^2x^2} \times 8a^3 x^6 = 80ax^4\\\\$