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4 years ago

When x2+3x-4 is subtracted from x (x2+3x-2) what is the difference

2 answers:

5 0

I'm assuming this is x^2 + 3x - 4 and x(x^2 + 3x - 2)

1.) First distribute x(x^2 + 3x - 2) to get x^3 + 3x^2 - 2x.

2.) Because you are subtracting all the terms from x^3 + 3x^2 - 2x, it's the same thing as distributing -1 to x^2 + 3x - 4 and then adding it to x^3 + 3x^2 - 2x.

3.) -1(x^2 + 3x - 4) = -x^2 - 3x + 4

4.) Add (x^3 + 3x^2 - 2x) + (-x^2 - 3x + 4)

5.) x^3 + 2x^2 - 5x + 4 is your final answer.

anygoal [31]4 years ago

3 0

Assuming that the problem is (x^2+3x-4) - x(x^2+3x-2) then this would be simple.. simplifying XD
here is the work for you:
(x^2+3x-4) - x(x^2+3x-2)
(x^2+3x-4) - (x^3+3x^2-2x)
Final Answer: -x^3-2x^2+5x-4

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Which of the following are remote interior angles of <6? check all that apply

vlada-n [284]

Answer:

C. <3, E. <1

Step-by-step explanation:

A triangle has 3 vertices, so it has exactly 3 interior angles, one at each vertex.

A triangle has 2 exterior angles at each vertex, so a triangle has 6 exterior angles. Each exterior angle is adjacent to an interior angle. The interior angles that are not adjacent to an exterior angle are that exterior angle's remote interior angles.

<6 is an exterior angle of the triangle. <5 is the other exterior angle at that vertex. <2 is an interior angle of the triangle and is adjacent to <6, so <2 is not a remote interior angle to <6.

The other two interior angles of the triangle are <1 and <3.

<1 and <3 are interior angles that are not adjacent to <6, so they are the remote interior angles to <6.

Answer: <1, <3

8 0

4 years ago

The least common denominator of 3/2 and 2/3 is

larisa86 [58]

Answer:

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

The table below represents a linear function f(x) and the equation represents a function g(x):

9966 [12]

Answer:

<u>Use 2 points from he table to find the slope of f(x):</u>

(0, -1) and (1, 7)

<u>The slope is:</u>

m = (7 - (-1))/1 = 8

The y - intercept is b = -1 according to point (0, -1).

<u>The lines are:</u>

f(x) = 8x - 1
g(x) = 3x - 2

The f(x) has greater slope since 8 > 3 and greater y-intercept since -1 > - 2.

See above

8 0

3 years ago

Write the quadratic function in the form g(x)=a (x-h)2 +kThen, give the vertex of its graph. G(x) =2x2 +20x+49Writing in the for

Snezhnost [94]

The quadratic function given to us is:

$g(x)=2x^2+20x+49$

We are asked to find the vertex form of the function.

The general formula for the vertex form of a quadratic equation is:

$\begin{gathered} g(x)=a(x-h)^2+k \\ \text{where,} \\ (h,k)\text{ is the coordinate of the vertex} \end{gathered}$

In order to write the function in its vertex form, we need to perform a couple of operations on the function.

1. Add and subtract the square of the half of the coefficient of x to the function.

2. Factor out the function with its repeated roots and re-write the equation.

Now, let us solve.

1. Add and subtract the square of the half of the coefficient of x to the function.

$\begin{gathered} g(x)=2x^2+20x+49=2(x^2+10x+\frac{49}{2}) \\ \text{half of coefficient of x:} \\ \frac{10}{2}=5 \\ \text{square of the half of the coefficient of x:} \\ 5^2=25 \\ \\ \therefore g(x)=2(x^2+10x+25-25+\frac{49}{2}) \end{gathered}$

2. Factor out the function with its repeated roots and re-write the equation.

$\begin{gathered} g(x)=2(x^2+10x+25)-2(25+\frac{49}{2}) \\ re-\text{write the above function} \\ g(x)=2(x^2+10x+25)-1 \\ \text{Let us factorize this:} \\ g(x)=2(x+5)^2-1 \end{gathered}$

Therefore, we can conclude that the Equation and vertex of the equation is: