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KIM [24]
3 years ago
6

Multiply. (x−5)(3x+6) Express the answer in standard form.

Mathematics
2 answers:
sergey [27]3 years ago
6 0

Answer:

3x^{2} -9x-30

Step-by-step explanation:

(x - 5)(3x + 6)

3x^{2} +6x-15x-30\\3x^{2} -9x-30

vodomira [7]3 years ago
5 0

Answer:

3x² - 9x - 30

General Formulas and Concepts:

<u>Algebra I</u>

  • Expand by FOIL (First Outside Inside Last)
  • Combining Like Terms

Step-by-step explanation:

<u>Step 1: Define</u>

(x - 5)(3x + 6)

<u>Step 2: Expand</u>

  1. Expand [FOIL]:                    3x² + 6x - 15x - 30
  2. Combine like terms:           3x² - 9x - 30
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What best describes the asymptote of an exponential function of the form f(x)=b^x?
horrorfan [7]

It depends on the value of b


Exponential functions are defined for positive values of the base [ŧex] b [/tex]. Also, they're not defined if b=1, or at least let's say that this is a trivial case, since it is the constant function 1.


Case 0<b<1:

If the base sits between 0 and 1, the exponential function is constantly decreasing. The explanation is quite intuitive, assume for the sake of simpleness that b is rational. If it sits between 0 and 1, it means that it can be written as a fraction p/q, with p<q. So, if you give a large exponent to b, you obtain


\frac{p^x}{q^x},\quad p^x \ll q^x \text{ as } x \to \infty


On the other hand, if you consider negative exponents, you switch numerator and denominator and then raise to the same exponent the fraction q/p, which gets larger and larger.


So, if 0<b<1, we have


\lim_{x \to -\infty} b^x = \infty,\qquad \lim_{x \to \infty} b^x = 0


and thus 0 is a horizontal asymptote as x tends to (positive) infinity.


Case b>1:

This case is very similar, except all roles are inverted. Now you start with a fraction p/q where p>q. So, with positive, large exponents you get


\frac{p^x}{q^x},\quad p^x \gg q^x \text{ as } x \to \infty


And as before, negative exponents switch numerator and denominator, so the fraction becomes q/p and thus you have


\lim_{x \to -\infty} b^x = 0,\qquad \lim_{x \to \infty} b^x = \infty


So, again, 0 is a horizontal asymptote, but this time for x tending towards negative infinite.

6 0
3 years ago
Read 2 more answers
Find the slope and y-intercept of the line is parallel to y=-1/3x+5 and passes through the point (8,2)
BartSMP [9]

Answer:

I know the slope is is the same since it is parallel! Hope that helped

Step-by-step explanation:


4 0
3 years ago
A rectangle has its base on the x axis and its upper two vertices on the parabola y = 9 − x2. (a) Draw a graph of this problem.
gtnhenbr [62]

Answer:

  • (a) see attached
  • (b) width: 2x; height: y = 9-x²
  • (c) A=2x(9-x²) . . . 0 ≤ x ≤ 3
  • (d) dA/dx = -6x² +18; x=±√3
  • (e) 12√3 units²

Step-by-step explanation:

(a) The attachment shows the graph of the parabola in blue. It also shows an inscribed rectangle in black.

(b) The upper right point of the rectangle is shown in the attachment as (x, y). The dimension y is the height of the rectangle. The x-dimension is half the width of the rectangle, which is symmetrical about the y-axis. Hence the width is 2x.

(c) As with any rectangle, the area is the product of length and width:

... A = (2x)(9 -x²) . . . . . the attachment shows a graph of this

... A = -2x³ +18x . . . . . expanded form suitable for differentiation

A suitable domain for A is where both x and A are non-negative: 0 ≤ x ≤ 3.

(d) The derivative of A with respect to x is ...

... A' = -6x² +18

This is defined everywhere, so the critical values will be where A' = 0.

... 0 = -6x² +18

... 3 = x² . . . . . . . divide by -6, add 3

... √3 = x . . . . . . . -√3 is also a solution, but is not in the domain of A

(e) The rectangle will have its largest area where x=√3. That area is ...

... A = 2x(9 -x²) = 2√3(9 -(√3)²) = 2√3(6)

... A = 12√3 . . . . square units . . . . ≈ 20.785 units²

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