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prisoha [69]
3 years ago
6

Find a1 for the geometric series r=3 and s6=364

Mathematics
2 answers:
givi [52]3 years ago
7 0

Answer: 1

Step-by-step explanation:

I typed in one to my aplus question and it was correct so the answer is 1

Neko [114]3 years ago
4 0
The answer is -1. hope i helped.
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Identify the<br> characteristics<br> of exponential<br> functions.
Masteriza [31]

Answer:

The graph passes through the point (0,1)

The domain is all real numbers.

The range is y>0.

The graph is decreasing.

The graph is asymptotic to the x-axis as x approaches positive infinity.

(These are just some of them by the way)

Step-by-step explanation:

5 0
2 years ago
Will someone help me with this please
ANEK [815]

the first answer choice, associative property of addition

hope this helps. gl!

8 0
3 years ago
Read 2 more answers
Determine the equations of the vertical and horizontal asymptotes, if any, for y=x^3/(x-2)^4
djverab [1.8K]

Answer:

Option a)

Step-by-step explanation:

To get the vertical asymptotes of the function f(x) you must find the limit when x tends k of f(x). If this limit tends to infinity then x = k is a vertical asymptote of the function.

\lim_{x\to\\2}\frac{x^3}{(x-2)^4} \\\\\\lim_{x\to\\2}\frac{2^3}{(2-2)^4}\\\\\lim_{x\to\\2}\frac{2^3}{(0)^4} = \infty

Then. x = 2 it's a vertical asintota.

To obtain the horizontal asymptote of the function take the following limit:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4}

if \lim_{x \to \infty}\frac{x^3}{(x-2)^4} = b then y = b is horizontal asymptote

Then:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4} \\\\\\lim_{x \to \infty}\frac{1}{(\infty)} = 0

Therefore y = 0 is a horizontal asymptote of f(x).

Then the correct answer is the option a) x = 2, y = 0

3 0
3 years ago
Read 2 more answers
Please help with Algebra
mina [271]

Answer:

C) 6

Step-by-step explanation:

I converted the equation into slope-intercept form and then identified the y-intercept.

4x+2y=12\\\\4x-4x+2y=12-4x\\\\2y=12-4x\\\\\frac{2y=12-4x}{2}\\\\ y=6-2x\\\\y=-2x+6\\\\y=-2x+\boxed{6}

C should be the correct answer.

Hope this helps.

3 0
2 years ago
Read 2 more answers
URGENT! (Grades close tomorrow and turning this in will boost my grade from a B to an A in College Algebra. I can't figure these
zvonat [6]

Answer:

  24.  see below for a drawing with axis intercepts labeled

  30.  the plane is parallel to the y-z plane. It is "x=4". The only axis intercept is (4, 0, 0).

Step-by-step explanation:

24. The axis intercepts are fairly easily found. You can do it a couple of ways.

<u>first way</u>

  • Set other variables to zero and divide by the coefficient of the variable of interest.

For 3x +5y +15z = 15, the x-intercept can be found by setting y and z to zero. This gives ...

  3x = 15

so, dividing by 3 tells you the x-intercept is ...

  x = 15/3 = 5

Likewise, the y- and z-intercepts are, respectively, ...

  y = 15/5 = 3

  z = 15/15 = 1

So, the points where the plane crosses the axes are ...

  (5, 0, 0), (0, 3, 0), (0, 0, 1) . . . . axes intercepts

__

<u>second way</u>

  • Divide by the constant on the right, and express all coefficients as denominators

Doing that gives ...

\dfrac{3x+5y+15z}{15}=1\\\\\dfrac{x}{5}+\dfrac{y}{3}+\dfrac{z}{1}=1

The denominators are the corresponding intercepts. (This is called the "intercept form" of the equation of the plane.)

From the above equation, we see the axis intercepts are ...

  (5, 0, 0), (0, 3, 0), (0, 0, 1) . . . . same as above

Actually, the math for these two methods is virtually identical. The second way does it "all at once", so can take fewer steps. Effectively, you're doing the math of the "first way" and expressing the result in the denominator.

__

Please note that when a variable is missing, there is no intercept for that axis. That is, the plane is parallel to that axis. (In problem 30, for example, the y- and z-variables are missing. The plane x=4 is parallel to the y- and z-axes (the y-z plane).)

_____

In the first attached drawing, the x-axis is red, the y-axis is green, and the z-axis is blue. The labeling of the points is difficult to see, but the axes are a different (lighter) color behind the plane than in front of it. This plane forms a triangle in the first octant (x, y, z > 0).

In the second attached drawing, the x=0 plane is blue; the y=0 plane is green, and the z=0 plane is red. The plane of the equation is yellow. This figure better shows the triangle in the first octant.

3-D graph paper (isometric graph paper) grids are available with a web search. Such might make it easier to draw the planes you need to draw. (See the third attachment for one example.)

3 0
3 years ago
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