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3 years ago

2. Graph the function y=x^2+2x-3 . Draw a chart to show at least three ordered pairs that are on the graph.

1 answer:

6 0

You'd have to draw a table of x and y values first to be able to graph.
Step 1: Pick any number for x. Let's pick 2.
Step 2: Plug in 2 into the equation: y = 2^2 + 2(2) - 3
Step 3: Solve for y: 4 + 4 - 3 = 5
Step 4: So the point (ordered pair) would be (x, y) ⇒ (2, 5). You can plot that on the graph.

Follow these steps 2 more times to get at least three ordered pairs.

Make sure you have enough points to know where the graph curves, then you can connect the dots and now there you have a graph and chart :)

You might be interested in

A circular mirror is surrounded by a square metal frame. The radius of the mirror is 5x. The side length of the metal frame is 1

Aleks04 [339]

Answer:

$-25x^{2} (\pi -9)$

Step-by-step explanation:

The area of the metal frame is equal to the area of the square minus the area of the circle.

The area of the square is the side length squared:

$(15x)^{2}$

$=225x^{2}$

The area of the circle is $\pi r^{2}$ , with $r$ being the radius:

$\pi (5x)^{2}$

$=\pi *25x^{2}$

$=25\pi x^{2}$

Now find the area of the metal frame:

$225x^{2} -25\pi x^{2}$

$25x^{2} (9-\pi )$ Factor out the GCF, $25x^{2}$

$-25x^{2} (\pi -9)$ Factor out a $-1$

5 0

3 years ago

Get it wrong and istg ima report all ur answers and questions until you get banned. Not a joke for people who think it is funny

lana [24]

Answer:

A, B, C, E

Step-by-step explanation:

A rhombus is the parallelogram with perpendicular diagonals. All

Correct statements, reasons:

A - adjacent angles are supplementary
B - opposite sides are parallel
C - diagonals are perpendicular
E - opposite sides are congruent

Incorrect statements, reasons:

D - opposite angles are congruent, but not supplementary
F - diagonals are not congruent, it would be square not rhombus

7 0

3 years ago

Read 2 more answers

X+4 is prime. X2-9can be factored using the __formula

svet-max [94.6K]

$\purple \bold{a^2 - b^2}$

Step-by-step explanation:

X+4 is prime. X2-9can be factored using the $\purple \bold{a^2 - b^2}$ formula

$x^2 - 9\\=x^2 - 3^2 \\= (x+3)(x-3)$

3 0

3 years ago

53.1 L =___ mL What is the answer to this?

Sveta_85 [38]

1 liter = 1000 ml
53.1 liter = 53.1*1000 = 53100

Hope this helps :)

6 0

3 years ago

Read 2 more answers

How would I solve 16x^4-41x^2+25=0 ???

sveticcg [70]

$16{ x }^{ 4 }-41{ x }^{ 2 }+25=0$

${ x }^{ 4 }={ ({ x }^{ 2 }) }^{ 2 }\\ \\ 16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0$

First of all to make our equation simpler, we'll equal $x^{2}$ to a variable like 'a'.

So,

${ x }^{ 2 }=a$

Now let's plug $x^{2}$ 's value (a) into the equation.

$16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0\\ \\ { x }^{ 2 }=a\\ \\ 16{ (a) }^{ 2 }-41{ a }+25=0$

Now we turned our equation into a quadratic equation.

(The variable 'a' will have a solution set of two solutions, but 'x' , which is the variable of our first equation will have a solution set of four solutions since it is a quartic equation (fourth-degree equation) )

Let's solve for a.

The formula used to solve quadratic equations ;

$\frac { -b\pm \sqrt { { b }^{ 2 }-4\cdot t\cdot c } }{ 2\cdot t }$

The formula is used in an equation formed like this :

$t{ x }^{ 2 }+bx+c=0$

In our equation,

t=16 , b=-41 and c=25

Let's plug the values in the formula to solve.

$t=16\quad b=-41\quad c=25\\ \\ \frac { -(-41)\pm \sqrt { -(41)^{ 2 }-4\cdot 16\cdot 25 } }{ 2\cdot 16 } \\ \\ \frac { 41\pm \sqrt { 1681-1600 } }{ 32 } \\ \\ \frac { 41\pm \sqrt { 81 } }{ 32 } \\ \\ \frac { 41\pm 9 }{ 32 }$

So the solution set :

$\frac { 41+9 }{ 32 } =\frac { 50 }{ 32 } \\ \\ \frac { 41-9 }{ 32 } =\frac { 32 }{ 32 } =1\\ \\ a\quad =\quad \left\{ \frac { 50 }{ 32 } ,\quad 1 \right\}$

We found a's value.

Remember,

${ x }^{ 2 }=a$

So after we found a's solution set, that means.

${ x }^{ 2 }=\frac { 50 }{ 32 }$

and

${ x }^{ 2 }=1$

We'll also solve this equations to find x's solution set :)

${ x }^{ 2 }=\frac { 50 }{ 32 } \\ \\ \frac { 50 }{ 32 } =\frac { 25 }{ 16 } \\ \\ { x }^{ 2 }=\frac { 25 }{ 16 } \\ \\ \sqrt { { x }^{ 2 } } =\sqrt { \frac { 25 }{ 16 } } \\ \\ x=\quad \pm \frac { 5 }{ 4 }$

${ x }^{ 2 }=1\\ \\ \sqrt { { x }^{ 2 } } =\sqrt { 1 } \\ \\ x=\quad \pm 1$

So the values x has are :

$\frac { 5 }{ 4 }$ , $-\frac { 5 }{ 4 }$ , $1$ and $-1$

Solution set :

$x=\quad \left\{ \frac { 5 }{ 4 } \quad ,\quad -\frac { 5 }{ 4 } \quad ,\quad 1\quad ,\quad -1 \right\}$

I hope this was clear enough. If not please ask :)

3 0

3 years ago

Read 2 more answers