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3 years ago

(20x^2 – 3x – 9)/ (4x – 3) Using synthetic division

Mathematics

2 answers:

Bond [772]3 years ago

4 0

Answer:

2x^2 + 2x +3

I think that this is correct

tigry1 [53]3 years ago

4 0

The answer above is t correct answer

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1.<br> Work<br> 8m + 5 = 6m + 17<br> Reason<br> Given Statement

snow_tiger [21]

Answer:

m = 6

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Writing equations of lines

Valentin [98]

Step-by-step explanation:

Equation of straight line is y=mx+c

choose any two points on straight line

for me I choose:(-3,11) and (3,-1)

use these two points to find gradient,m.

m= (-1-11)/(3-(-3))

m= -2

now, y=-2x+c

choose any point on the straight line

I choose point (3,-1)

sub the point into the equation to find c

-1=-2(3)+c

c=5

equation: y=-2x+5

7 0

3 years ago

Let f (x) = 3x − 1 and ε > 0. Find a δ > 0 such that 0 < ∣x − 5∣ < δ implies ∣f (x) − 14∣ < ε. (Find the largest

s344n2d4d5 [400]

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and $\varepsilon>0$ , we want to find a $\delta$ such that $|f(x)-14|$

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that $|x-5|$ . Then

$|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta$ .

Then, if $|x-5|$ , then $|f(x)-14|$ . So, in this case, if $3\delta \leq \varepsilon$ we get that $|f(x)-14|$ . The maximum value of delta is $\frac{\varepsilon}{3}$ .

By definition, this procedure proves that $\lim_{x\to 5}f(x) = 14$ . Note that f(5)=14, so this proves that f is continous at x=5.

3 0

2 years ago

The quadratic formula gives which roots for the equation 2x^2 + x - 6 = 0?

Jet001 [13]

Answer:

The roots are $x = (\frac{3}{2}, -2)$ , which is given by option C.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

In this question, we have that:

$2x^2 + x - 6 = 0$

Which is a quadratic equation with $a = 2, b = 1, c = -6$ . So

$\Delta = 1^{2} - 4*2(-6) = 1 + 48 = 49$

$x_{1} = \frac{-1 + \sqrt{49}}{2*2} = \frac{6}{4} = \frac{3}{2}$

$x_{2} = \frac{-1 - \sqrt{49}}{2*2} = \frac{-8}{4} = -2$

So the roots are $x = (\frac{3}{2}, -2)$ , which is given by option C.

3 0

3 years ago

PLEASE HELP AGAIN ASAP!

Dovator [93]

15/5 goes to all of them..

15/5 = 3

8 0

3 years ago