How many kilograms are equal to 907.1 grams.

Factor each.<br><br><br> Please help!

Naddika [18.5K]

Answer:

1) The factors of $x^2-3x+2=0$ are $\mathbf{(x-1)(x-2)=0}$

Option C is correct.

3) The factors of $x^2+2x-8=0$ are $\mathbf{(x-2)(x+4)=0}$

Option B is correct.

Step-by-step explanation:

1) Factor : $x^2-3x+2=0$

For factoring we need to break the middle term, such that there sum is equal to middle term of expression and product is equal to product of first and last term.

The middle term : -3x

We can break them as (-2x)( -x)

Solving:

$x^2-3x+2=0\\x^2-2x-x+2=0\\x(x-2)-1(x-2)=0\\(x-1)(x-2)=0$

So, factors of $x^2-3x+2=0$ are $\mathbf{(x-1)(x-2)=0}$

Option C is correct.

3) Factor: $x^2+2x-8=0$

For factoring we need to break the middle term, such that there sum is equal to middle term of expression and product is equal to product of first and last term.

The middle term : 2x

We can break them as (4x)( -2x)

Solving

$x^2+2x-8=0\\x^2+4x-2x-8=0\\x(x+4)-2(x+4)=0\\(x-2)(x+4)=0$

So, factors of $x^2+2x-8=0$ are $\mathbf{(x-2)(x+4)=0}$

Option B is correct.

7 0

3 years ago

Help please i need it really bad pls :(

Luden [163]

Answer:

option 2

hope this helps

have a good day :)

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Graph the functions on the same coordinate axis. {f(x)=−2x+1g(x)=x2−2x−3

NikAS [45]

Answer:

(2,-3) and (-2,5)

Step-by-step explanation:

Let us graph the two equations one by one.

1. $f(x)=-2x+1$

If we compare this equation with the slope intercept form of a line which is given as

$y=mx+c$

we see that m = -1 and c =1

Hence the slope of the line is -2 and the y intercept is 1. Hence one point through which it is passing is (0,1) .

Let us find another point by putting x = 1 and solving it for y

$y=-2(1)+1$

$y=-2+1 = -1$

Let us find another point by putting x = 2 and solving it for y

$y=-2(2)+1$

$y=-4+1 = -3$

Hence the another point will be (2,-3)

Let us find another point by putting x = -2 and solving it for y

$y=-2(-2)+1$

$y=+1 = 5$

Hence the another point will be (-2,5)

Now we have two points (0,1) ,(1,-1) , (2,-3) and (-2,5) we joint them on line to obtain our line

2.

$g(x)=y=x^2-2x-3$

$y=x^2-2x+1-1-3$

$y=(x-1)^2-4$

$(y+4)=(x-1)^2$

It represents the parabola opening upward with vertices (1,-4)

Let us mark few coordinates so that we may graph the parabola.

i) x=0 ; $y=y=(0)^2-2(0)-3=0-0-3=-3$ ; (0,-3)

ii)x=-1 ; $y=(-1)^2-2(-1)-3=1+2-3=0$ ; (-1,0)

iii) x=2 ; $y=(2)^2-2(2)-3 = 4-4-3 =-3$ ;(2,-3)

iii) x=1 ; $y=(1)^2-2(1)-3 = 1-2-3 =-4$ ;(1,-4)

iii) x=-2 ; $y=(-2)^2-2(-2)-3 = 4+4-3 =5$ ;(-2,5)

Now we plot them on coordinate axis and line them to form our parabola

When we plot them we see that we have two coordinates (2,-3) and (-2,5) are common , on which our graphs are intersecting. These coordinates are solution to the two graphs.