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4 years ago

As x increases without bound, f(x):

Mathematics

1 answer:

yarga [219]4 years ago

7 0

Answer: Choice D) approaches y = -4

"x increases without bound" is another way of saying "x heads off to positive infinity". Visually, you follow the graph curve going to the right. As the graph shows, the curve steadily gets closer to the horizontal line y = -4, but it never actually gets there. This line is the horizontal asymptote.

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In a box there are six envelopes each containing two cards. Three of the envelopes contain two red cards, two of them contain a

Sphinxa [80]

Answer:

$\frac{2}{5}$

Step-by-step explanation:

3 envelopes having 2 red card

2 envelopes having 1 red card and 1 black card

1 envelope having 2 black cards

We are given that . An envelope is selected at random and a card is withdrawn and found to be red.

So, No. of ways of envelope having red card = 3+2 = 5

No. of required ways of envelope having 1 red card and 1 black card = 2

So, probability of getting an envelope having 1 red card and 1 black card = $\frac{2}{5}$

Hence The chance the other card is black is $\frac{2}{5}$

5 0

3 years ago

Complete the equation, by supplying the missing exponent.<br> m•n2•m3= m 11n2

MaRussiya [10]

Given:

Consider the given equation is:

$m^?\cdot n^2\cdot m^3=m^{11}\cdot n^2$

To find:

The missing exponent.

Solution:

Let x be the missing exponent. Then the given equation can be written as

$m^x\cdot n^2\cdot m^3=m^{11}\cdot n^2$

It can be rewritten as:

$(m^x\cdot m^3)\cdot n^2=m^{11}\cdot n^2$

$m^{x+3}\cdot n^2=m^{11}\cdot n^2$ $[\because a^ma^n=a^{m+n}]$

On comparing the coefficient of m, we get

$x+3=11$

$x=11-3$

$x=8$

Therefore, the value of the missing exponent is 8. So, the complete equation is $m^8\cdot n^2\cdot m^3=m^{11}\cdot n^2$ .

4 0

3 years ago

Enter a numerical expression for the word expression 10 decreased by 4

bulgar [2K]

10-4=6. can it be like this.

7 0

3 years ago

Read 2 more answers

Which function is the same as y = 3 cosine (2 (x startfraction pi over 2 endfraction)) minus 2? y = 3 sine (2 (x startfraction p

kirza4 [7]

The function which is same as the function y = 3cos(2(x +π/2)) -2 is: Option A: y= 3sin(2(x + π/4)) - 2

<h3>How to convert sine of an angle to some angle of cosine?</h3>

We can use the fact that:

$\sin(\theta) = \cos(\pi/2 - \theta)\\\sin(\theta + \pi/2) = -\cos(\theta)\\\cos(\theta + \pi/2) = \sin(\theta)$

to convert the sine to cosine.

<h3>Which trigonometric functions are positive in which quadrant?</h3>

In first quadrant (0 < θ < π/2), all six trigonometric functions are positive.
In second quadrant(π/2 < θ < π), only sin and cosec are positive.
In the third quadrant (π < θ < 3π/2), only tangent and cotangent are positive.
In fourth (3π/2 < θ < 2π = 0), only cos and sec are positive.

(this all positive negative refers to the fact that if you use given angle as input to these functions, then what sign will these functions will evaluate based on in which quadrant does the given angle lies.)

Here, the given function is:

$y= 3\cos(2(x + \pi/2)) - 2$

The options are:

$y= 3\sin(2(x + \pi/4)) - 2$
$y= -3\sin(2(x + \pi/4)) - 2$
$y= 3\cos(2(x + \pi/4)) - 2$
$y= -3\cos(2(x + \pi/2)) - 2$

Checking all the options one by one:

Option 1: $y= 3\sin(2(x + \pi/4)) - 2$

$y= 3\sin(2(x + \pi/4)) - 2\\y= 3\sin (2x + \pi/2) -2\\y = -3\cos(2x) -2\\y = 3\cos(2x + \pi) -2\\y = 3\cos(2(x+ \pi/2)) -2$

(the last second step was the use of the fact that cos flips its sign after pi radian increment in its input)
Thus, this option is same as the given function.

Option 2: $y= -3\sin(2(x + \pi/4)) - 2$

This option if would be true, then from option 1 and this option, we'd get:
$-3\sin(2(x + \pi/4)) - 2= -3\sin(2(x + \pi/4)) - 2\\2(3\sin(2(x + \pi/4))) = 0\\\sin(2(x + \pi/4) = 0$

which isn't true for all values of x.

Thus, this option is not same as the given function.

Option 3: $y= 3\cos(2(x + \pi/4)) - 2$

The given function is $y= 3\cos(2(x + \pi/2)) - 2 = 3\cos(2x + \pi) -2 = -3\cos(2x) -2$

This option's function simplifies as:

$y= 3\cos(2(x + \pi/4)) - 2 = 3\cos(2x + \pi/2) -2 = -3\sin(2x) - 2$

Thus, this option isn't true since $\sin(2x) \neq \cos(2x)$ always (they are equal for some values of x but not for all).

Option 4: $y= -3\cos(2(x + \pi/2)) - 2$

The given function simplifies to: $y= 3\cos(2(x + \pi/2)) - 2 = 3\cos(2x + \pi) -2 = -3\cos(2x) -2$

The given option simplifies to:

$y= -3\cos(2(x + \pi/2)) - 2 = -3\cos(2x + \pi ) -2\\y = 3\cos(2x) -2$

Thus, this function is not same as the given function.

Thus, the function which is same as the function y = 3cos(2(x +π/2)) -2 is: Option A: y= 3sin(2(x + π/4)) - 2

Learn more about sine to cosine conversion here:

brainly.com/question/1421592

4 0

2 years ago

Read 2 more answers

A population of 400 beetles grows by 5% each week. How can the beetle population be determined after a number of weeks, w?

Helga [31]

For this case we have an equation of the form:
f (w) = A * (b) ^ w
Where,
A: initial amount
b: growth rate
w: number of weeks
Substituting values we have:
f (w) = 400 * (1.05) ^ w
Answer:
the beetle population can be determined after a number of weeks, w, with the following function:
f (w) = 400 * (1.05) ^ w

6 0

3 years ago

Read 2 more answers