All categories

3 years ago

How to find the zeroes by factoring

Mathematics

1 answer:

Rashid [163]3 years ago

7 0

Your set the equation equal to 0 and you start factoring it, if it doesn’t factor out then you use the quadratic formula but if you do that then set it equal to “X”, not equal to 0

You might be interested in

Given below are the graphs of two lines, y=-0.5 + 5 and y=-1.25x + 8 and several regions and points are shown. Note that C is th

zalisa [80]

We have the following equations:

$(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8$

So we are asked to write a system of equations or inequalities for each region and each point.

Part a)

Region Example A

$y \leq -0.5x+5 \\ y \leq -1.25x+8$

Region B.

Let's take a point that is in this region, that is:

$P(0,6)$

So let's find out the signs of each inequality by substituting this point in them:

$y \ (?)-0.5x+5 \\ 6 \ (?) -0.5(0)+5 \\ 6 \ (?) \ 5 \\ 6\ \textgreater \ 5 \\ \\ y \ (?) \ -1.25x+8 \\ 6 \ (?) -1.25(0)+8 \\ 6 \ (?) \ 8 \\ 6\ \textless \ 8$

So the inequalities are:

$(1) \ y \geq -0.5x+5 \\ (2) \ y \leq -1.25x+8$

Region C.

A point in this region is:

$P(0,10)$

So let's find out the signs of each inequality by substituting this point in them:

$y \ (?)-0.5x+5 \\ 10 \ (?) -0.5(0)+5 \\ 10 \ (?) \ 5 \\ 10\ \textgreater \ 5 \\ \\ y \ (?) \ -1.25x+8 \\ 10 \ (?) -1.25(0)+8 \\ 10 \ (?) \ 8 \\ 10 \ \ \textgreater \ \ 8$

So the inequalities are:

$(1) \ y \geq -0.5x+5 \\ (2) \ y \geq -1.25x+8$

Region D.

A point in this region is:

$P(8,0)$

So let's find out the signs of each inequality by substituting this point in them:

$y \ (?)-0.5x+5 \\ 0 \ (?) -0.5(8)+5 \\ 0 \ (?) \ 1 \\ 0 \ \ \textless \ \ 1 \\ \\ y \ (?) \ -1.25x+8 \\ 0 \ (?) -1.25(8)+8 \\ 0 \ (?) \ -2 \\ 0 \ \ \textgreater \ \ -2$

So the inequalities are:

$(1) \ y \leq -0.5x+5 \\ (2) \ y \geq -1.25x+8$

Point P:

This point is the intersection of the two lines. So let's solve the system of equations:

$(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8 \\ \\ Subtracting \ these \ equations: \\ 0=0.75x-3 \\ \\ Solving \ for \ x: \\ x=4 \\ \\ Solving \ for \ y: \\ y=-0.5(4)+5=3$

Accordingly, the point is:

$\boxed{p(4,3)}$

Point q:

This point is the $x-intercept$ of the line:

$y=-0.5x+5$

So let:

$y=0$

Then

$x=\frac{5}{0.5}=10$

Therefore, the point is:

$\boxed{q(10,0)}$

Part b)

The coordinate of a point within a region must satisfy the corresponding system of inequalities. For each region we have taken a point to build up our inequalities. Now we will take other points and prove that these are the correct regions.

Region Example A

The origin is part of this region, therefore let's take the point:

$O(0,0)$

Substituting in the inequalities:

$y \leq -0.5x+5 \\ 0 \leq -0.5(0)+5 \\ \boxed{0 \leq 5} \\ \\ y \leq -1.25x+8 \\ 0 \leq -1.25(0)+8 \\ \boxed{0 \leq 8}$

It is true.

Region B.

Let's take a point that is in this region, that is:

$P(0,7)$

Substituting in the inequalities:

$y \geq -0.5x+5 \\ 7 \geq -0.5(0)+5 \\ \boxed{7 \geq \ 5} \\ \\ y \leq \ -1.25x+8 \\ 7 \ \leq -1.25(0)+8 \\ \boxed{7 \leq \ 8}$

It is true

Region C.

Let's take a point that is in this region, that is:

$P(0,11)$

Substituting in the inequalities:

$y \geq -0.5x+5 \\ 11 \geq -0.5(0)+5 \\ \boxed{11 \geq \ 5} \\ \\ y \geq \ -1.25x+8 \\ 11 \ \geq -1.25(0)+8 \\ \boxed{11 \geq \ 8}$

It is true

Region D.

Let's take a point that is in this region, that is:

$P(9,0)$

Substituting in the inequalities:

$y \leq -0.5x+5 \\ 0 \leq -0.5(9)+5 \\ \boxed{0 \leq \ 0.5} \\ \\ y \geq \ -1.25x+8 \\ 0 \geq -1.25(9)+8 \\ \boxed{0 \geq \ -3.25}$

It is true

7 0

4 years ago

Help fast pls

IgorC [24]

Answer:

x=2

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

If sin 0=7/25 find csc 0

sukhopar [10]

Answer:

csc theta = 25/7

Step-by-step explanation:

I believe you meant the Greek character theta, not 0. Note that sin 0 = 0, which is definitely not equal to 7/25.

Rewriting the given info, we obtain:

sin theta = 7/25 find csc theta

The csc function is the reciprocal of the sin function. Therefore,

if sin theta = 7/25, then csc theta = 25/7

7 0

3 years ago

The area of a square is 361 square feet. What is the distance around the square?

Ray Of Light [21]

The area of a square is its length times its width: l x w = 361
Since the length of a square is the same as its width, replace 'w' with 'l': l x l = 361
l x l = l² = 361 ---> l = √361 ---> l = √361 ---> l = 19 ft.

3 0

3 years ago

Read 2 more answers

What is the value of x?

uranmaximum [27]

x = 41 cm

Here is why:

For this question we have to use the pythagorean theorem, which is used for right triangles. The hypotenuse will always be the last in this theorem.

$a^2+b^2=c^2$

Plug in.

$9^2+40^2=c^2$

$81+1,600=c^2$

Add together.

$1,681=c^2$

Now we will find the square root of 1,681, since c is squared.

$\sqrt{1,681}$

= 41 cm

5 0

3 years ago