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3 years ago

SOMEONE PLEASE HELP ME I’m really confused

Mathematics

1 answer:

SOVA2 [1]3 years ago

6 0

Answer:

Were you given a graph or do you have one you can use?

- If you do, all the need to do is plot the points that are shown in the question. -Secondly, You need to join the lines together.

-After that, you need to find the 'middle point' of each line. Eg. If I had a line with 4 spaces in-between it, I would put a point or cross after the 2nd space.

Hope that helped? Sorry if it's too confusing and was useless.

In the image the blue shape is the rectangle and the red shape is the rhombus/tilted square. (MY DRAWING IS BAD I KNOW DON'T SAY ANYTHING ABOUT IT) Anyways. I still feel like you need to do it yourself as this was assigned to you so get some paper and draw this graph out. Let me know if you need more help!

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Salsk061 [2.6K]

Given that the two functions are $f(x)=9x+2$ and $g(x)=-9x^2-2x+1$

We need to determine the value of $(f \circ g)(-6)$

The value of $(f \circ g)(x)$ :

The value of $(f \circ g)(x)$ can be determined using the formula,

$(f \circ g)(x)=f[g(x)]$

Substituting $g(x)=-9x^2-2x+1$ in the above formula, we get;

$(f \circ g)(x)=f[-9x^2-2x+1]$

Now, substituting $x=-9 x^{2}-2 x+1$ in the function $f(x)=9x+2$ , we get;

$(f \circ g)(x)=9(-9x^2-2x+1)+2$

$(f \circ g)(x)=-81x^2-18x+9+2$

$(f \circ g)(x)=-81x^2-18x+11$

Thus, the value of $(f \circ g)(x)$ is $(f \circ g)(x)=-81x^2-18x+11$

The value of $(f \circ g)(-6)$ :

The value of $(f \circ g)(-6)$ can be determined by substituting x = -6 in the function $(f \circ g)(x)=-81x^2-18x+11$

Thus, we have;

$(f \circ g)(-6)=-81(-6)^2-18(-6)+11$

$(f \circ g)(-6)=-81(36)-18(-6)+11$

$(f \circ g)(-6)=-2916+108+11$

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Hence, Option B is the correct answer.

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$16y^5 + 4y^5 -12y^3 + 15y^3$

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$20y^5+3y^3$

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Answer:

$a'(d) = \frac{d}{5} + \frac{3}{5}$

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Step-by-step explanation:

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Solving (a): Write as inverse function

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Swap positions of d and y

$d = 5y - 3$

Make y the subject

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Replace y with a'(d)

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Prove that a(d) and a'(d) are inverse functions

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To do this, we prove that:

$a(a'(d)) = a'(a(d)) = d$

Solving for $a(a'(d))$

$a(a'(d)) = a(\frac{d}{5} + \frac{3}{5})$

Substitute $\frac{d}{5} + \frac{3}{5}$ for d in $a(d) = 5d - 3$

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$a(a'(d)) = \frac{5d}{5} + \frac{15}{5} - 3$

$a(a'(d)) = d + 3 - 3$

$a(a'(d)) = d$

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$a'(a(d)) = a'(5d - 3)$

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Add fractions

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