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3 years ago

5

Solving equations -2(x + 3)=3(2x-10)

1 answer:

nika2105 [10]3 years ago

3 0

Answer:

x=3

Step-by-step explanation:

You might be interested in

A bag contains 11 red chips and 12 blue chips. Two chips are selected randomly without replacement from the bag. What is the pro

Elena-2011 [213]

Answer:

11/23 or 0.478

Step-by-step explanation:

1. Find the probability of selecting 2 red chips:

11/23 * 10/22 = 5/23

2. Find the probability of selecting 2 blue chips:

12/23 * 11/22 = 6/23

3. Add the two probabilities:

5/23 + 6/23 = 11/23 = 0.478 (rounded to 3 decimal places)

hope this helps! <3

4 0

2 years ago

HELP WITH ALL <br> Determine the composite functions of the following set of functions

elixir [45]

QUESTION 1

i) Given $f(x)=5x-2$ and $g(x)=x^2$ .

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

$(f\circ g)=f(x^2)$

We plug in $x^2$ into $f(x)=5x-2$

$(f\circ g)=5x^2-2$

ii) Given $f(x)=5x-2$ and $g(x)=x^2$ .

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

$(g\circ f)=g(5x-2)$

We plug in $5x-2$ into $g(x)=x^2$

$(g\circ f)=(5x-2)^2$

$(g\circ f)=25x^2-10x-10x+4$

$(g\circ f)=25x^2-20x+4$

QUESTION 2

i) Given $g(x)=x^2-4$ and $h(x)=\sqrt{2x+1}$ .

$(h\circ g)=h(g(x))$

$(h\circ g)=h(x^2-4)$

We plug in $x^2-4$ into $h(x)=\sqrt{2x+1}$

$(h\circ g)=\sqrt{2(x^2-4)+1}$

$(h\circ g)=\sqrt{2x^2-8+1}$

$(h\circ g)=\sqrt{2x^2-7}$

ii) Given $g(x)=x^2-4$ and $h(x)=\sqrt{2x+1}$ .

$(g\circ h)=g(h(x))$

$(g\circ h)=g(\sqrt{2x+1})$

We plug in $\sqrt{2x+1$ into $g(x)=x^2-4$

$(g\circ h)=(\sqrt{2x+1})^2-4$

$(g\circ h)=2x+1-4$

$(g\circ h)=2x-3$

QUESTION 3

i) Given $g(x)=x+5$ and $h(x)=\sqrt{x-4}$ .

$(g\circ h)=g(h(x))$

$(g\circ h)=g(\sqrt{x-4})$

We plug in $\sqrt{x-4}$ into $g(x)=x+5$

$(g\circ h)=\sqrt{(x-4)+5}$

ii) Given $g(x)=x+5$ and $h(x)=\sqrt{x-4}$ .

$(g\circ g)=g(g(x))$

$(g\circ g)=g(x+5)$

We plug in $x+5$ into $g(x)=x+5$

$(g\circ g)=(x+5)+5}$

$(g\circ g)=x+10}$

QUESTION 4

The given functions are;

$f(x)=-2x+1$ and $g(x)=-x^2$

$f\circ f=f(f(x))$

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

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

$f\circ f=4x-2+1)$

$f\circ f=4x-1)$

ii)

$g\circ g=g(g(x))$

$g\circ g=g(-x^2)$

$g\circ g=-(x^2)^2$

$g\circ g=-x^4$

QUESTION 5

$f(x)=5x-2;g(x)=x^2;h(x)=-3x$

$h\circ g\circ f=h\circ (g(f(x))$

$h\circ g\circ f=h\circ (g(5x-2))$

$h\circ g\circ f=h\circ ((5x-2)^2)$

$h\circ g\circ f=h\circ (25x^2-20x+4)$

$h\circ g\circ f=h(25x^2-20x+4)$

$h\circ g\circ f=-3(25x^2-20x+4)$

$h\circ g\circ f=-75x^2+60x-12)$

ii.

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

$f\circ f\circ g=f\circ (f(x^2)$

$f\circ f\circ g=f\circ (5x^2-2)$

$f\circ f\circ g=f(5x^2-2)$

$f\circ f\circ g=5(5x^2-2)-2$

$f\circ f\circ g=25x^2-10-2$

$f\circ f\circ g=25x^2-12$

7 0

3 years ago

-4x -2 + 3x - 10 How many terms are there? SIMPLIFY. What is the coefficient of x?

Maru [420]

Answer:

Number of coefficients of x: 2 coefficients (-4 and 3)

Number of terms: They're 4 terms. (-4x, -2, + 3x, and -10)

Simplified: - x - 12

Step-by-step explanation:

1) Subtract the numbers.

-4x - 12 + 3x

2) Combine like terms.

- x - 12

6 0

2 years ago

Find AM in the parallelogram if PN =15 and AO = 6. The diagram is not to scale.

valentinak56 [21]

Answer:

6

Step-by-step explanation:

The diagonals of a parallelogram bisect each other.

The diagonals are PN and MO. The point A divides PN into 2 equal segments as well as MO into 2 equal segments.

Since AO = 6, MA should also be equal. Hence, MA (or AM) = 6.

The correct answer is AM = 6

7 0

3 years ago

Part 3: Find the inverse function, f –1(x), for the function f (x) = x2 + 10x + 28.

jasenka [17]

Answer:

-5+(√x-3) and -(√x-3)-5

Step-by-step explanation:

change f(x) to 'y' and then interchange x and y; then solve for y

y = x² + 10x + 28

x = y² + 10y + 28

you can use the Quadratic Formula to evaluate y² + 10y + 28 and will get:

-5+(√x-3) and -(√x-3)-5

6 0

3 years ago