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2 years ago

Find the single number that will make these two expressions equivalent.

Mathematics

1 answer:

Triss [41]2 years ago

7 0

Answer:

Which two expressions?

Step-by-step explanation:

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What is the range of the function graphed below?

lord [1]

Answer:

__________-3<Y<4______________

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2 years ago

If fog(x)=2x-1/x and g(x)=5x+2 , find f(x).

Vikentia [17]

Answer:

Step-by-step explanation:

To do this, we will use a u substitution. Namely, let 5x + 2 = u and solve for x so u is in terms of x, since we have x in our composition.

5x + 2 = u, then

5x = u - 2 and

$x=\frac{u-2}{5}$

which changes f(g(x)) into f(u) {since g(x) = 5x + 2 = u}:

$f(u)=\frac{2(\frac{u-2}{5})-1 }{\frac{u-2}{5} }$ and of course that mess needs to be simplified:

$f(u)=\frac{\frac{2u-4}{5}-1 }{\frac{u-2}{5} }$ and

$f(u)=\frac{\frac{2u-4}{5}-\frac{5}{5} }{\frac{u-2}{5} }$ and

$f(u)=\frac{\frac{2u-4-5}{5} }{\frac{u-2}{5} }$ and

$f(u)=\frac{\frac{2u-9}{5} }{\frac{u-2}{5} }$ and bring up the lower fraction and flip it to multiply to get

$f(u)=\frac{2u-9}{u-2}$ and now we can change the u to an x to get that

$f(x)=\frac{2x-9}{x-2}$

8 0

2 years ago

Question 9<br> The triangles shown are

yaroslaw [1]

Answer:

The triangles shown are<u> proportional</u>

Step-by-step explanation:

Two parallel lines are cut by a tranversal so the opposite and corresponding angles are congruent.

B = N

M = M

12 x 2 = 24 , So the sides of the triangle are proportional meaning they are parallel.

Side MN would be 12

Side MP would be 16

6 0

2 years ago

Somebody help.. im so lost!!

aleksley [76]

What type of math is this?

4 0

2 years ago

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

7 0

3 years ago