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slavikrds [6]
3 years ago
12

Write each of the following products in standard polynomial form. (a) (x+3)(x-2)(x-8) (b) (x+2)(x-2)(x+3)(x-3)

Mathematics
1 answer:
Grace [21]3 years ago
6 0
Each of the following products
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Please help! It’s due tomorrow it also says to show my work
Pani-rosa [81]

Given:

The inequality is

x+19\leq 23

To find:

The value of x and then graph on the number line.

Solution:

We have,

x+19\leq 23

Subtracting 19 from both sides, we get

x+19-19\leq 23-19

x\leq 4

It means all the real values of x which are less than or equal to 4 are in the solution set. So, an arrow approaches towards left from x=4 as shown in the below figure.

6 0
2 years ago
Question 3(Multiple Choice Worth 4 points) (05.05)The width of a rectangle is shown below: A coordinate plane with a point A at
olganol [36]
D 4/147
FTC +65-/325
Junk. 5(-4)
8 0
3 years ago
What is 8/12 in simplest form
Pani-rosa [81]
8/12 in simplest form is 2/3.
5 0
3 years ago
(10.02)
ollegr [7]
<h2>Hello!</h2>

The answer is:

C. Cosine is negative in Quadrant III

<h2>Why?</h2>

Let's discard each given option in order to find the correct:

A. Tangent is negative in Quadrant I: It's false, all functions are positive in Quadrant I (0° to 90°).

B. Sine is negative in Quadrant II: It's false, sine is negative in positive in Quadrant II. Sine function is always positive coming from 90° to 180°.

C. Cosine is negative in Quadrant III. It's true, cosine and sine functions are negative in Quadrant III (180° to 270°), meaning that only tangent and cotangent functions will be positive in Quadrant III.

D. Sine is positive in Quadrant IV: It's false, sine is negative in Quadrant IV. Only cosine and secant functions are positive in Quadrant IV (270° to 360°)

Have a nice day!

6 0
3 years ago
What is the equation of the line, in standard form, that passes through (4, -3) and is parallel to the line whose equation is 4x
Serggg [28]

Answer:

The equation of this line would be 4x + y = 13

Step-by-step explanation:

In order to find this equation we must first find the slope of the original line. To do this, we solve the original equation for y.

4x + y - 2 = 0

4x + y = 2

y = -4x + 2

The original slope (the coefficient of x) is -4, which means the new slope will also be -4 because parallel lines have the same slope. Now, we can use this slope along with the point in point-slope form to find the equation of the line. Just plug in the numbers and solve for the coefficient.

y - y1 = m(x - x1)

y + 3 = -4(x - 4)

y + 3 = -4x + 16

4x + y + 3 = 16

4x + y = 13

4 0
3 years ago
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