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2 years ago

Verify if this is an identity cot(-x) tanx=-1

1 answer:

7 0

<span>hello :
</span>cot(-x) tan(x)=-1......(*)
* cot(x) = 1/tan(x)
cot(-x) = -cot(x)= -(1/tan(x))

cot(-x) tan(x) = -(1/tan(x))×tan(x)=-1

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Honda Atlas issued 100,000 shares of stock. In 1990, each share of stock was worth Rs. 152.50. In 1993, each share of the stock

noname [10]

Answer:

Rs.13,130 ,000

Step-by-step explanation:

Lets solve first for year 1990

Value of stock in 1990 = Rs. 152.50

No. of stocks issued in 1990 = 100,000

Total value of these stocks in 1990= Value of stock in 1990* no of stocks issued in 1990 = 152.50*100,000 = Rs. 15,250,000

Let solve first for year 1993

Value of stock in 1993 = Rs. 21.20

No. of stocks 1993 = 100,000

Total value of these stocks in 1993 = Value of stock in 1993* no of stocks issued in 1993 = 21.20*100,000 = Rs. 2,120,000

Decrease in value of stock in 1993 in comparison to 1990 = Total value of the stocks in 1990 -Total value of the stocks in 1993

= Rs. 15,250,000 - Rs. 2,120,000

= Rs.13,130 ,000

Thus, the 100,000 shares worth Rs.13,130 ,000 lesser in 1993 than in 1990.

6 0

3 years ago

Where do you find a vertical shift in the exponential parent function?

Montano1993 [528]

For the given expressions we will have:

y = exp(x - 4) →we have a shift of 4 units to the right.

y = exp (x +9) → we have a shift of 9 units to the left.

y = exp(x) + 7 → we have a shift of 7 units up.

y = exp(x) - 6 → we have a shift of 6 units down.

<h3>How to work with vertical and horizontal shifts?</h3>

Remember that the shifts work as follows.

For a function f(x), we define a vertical shift of N units as:

g(x) = f(x) + N

If N > 0, the shift is upwards.
If N < 0, the shift is downwards.

For a function f(x), we define a horizontal shift of N units as:

g(x) = f(x + N)

If N > 0, the shift is to the left.
If N < 0, the shift is to the right.

Then, if we have:

exp(x - 4) we have a shift of 4 units to the right.

exp (x +9) we have a shift of 9 units to the left.

exp(x) + 7 we have a shift of 7 units up.

exp(x) - 6 we have a shift of 6 units down.

Learn more about translations

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4 0

1 year ago

Find the area between the graph of the function and the x-axis over the given interval, if possible.

vodomira [7]

Answer:

$A= -\frac{5}{-1} - \lim_{x\to\infty} \frac{5}{x-2} = 5-0 = 5$

So then the integral converges and the area below the curve and the x axis would be 5.

Step-by-step explanation:

In order to calculate the area between the function and the x axis we need to solve the following integral:

$A = \int_{-\infty}^1 \frac{5}{(x-2)^2}$

For this case we can use the following substitution $u = x-2$ and we have $dx = du$

$A = \int_{a}^b \frac{5}{u^2} du = 5\int_{a}^b u^{-2}du$

And if we solve the integral we got:

$A= -\frac{5}{u} \Big|_a^b$

And we can rewrite the expression again in terms of x and we got:

$A = -\frac{5}{x-2} \Big|_{-\infty}^1$

And we can solve this using the fundamental theorem of calculus like this:

$A= -\frac{5}{-1} - \lim_{x\to\infty} \frac{5}{x-2} = 5-0 = 5$

So then the integral converges and the area below the curve and the x axis would be 5.

7 0

3 years ago

I need help ASAP! Can anyone please check my work?

STALIN [3.7K]

A = event the person got the class they wanted

B = event the person is on the honor roll

P(A) = (number who got the class they wanted)/(number total)

P(A) = 379/500

P(A) = 0.758

There's a 75.8% chance someone will get the class they want

Let's see if being on the honor roll changes the probability we just found

So we want to compute P(A | B). If it is equal to P(A), then being on the honor roll does not change P(A).

---------------

A and B = someone got the class they want and they're on the honor roll

P(A and B) = 64/500

P(A and B) = 0.128

P(B) = 144/500

P(B) = 0.288

P(A | B) = P(A and B)/P(B)

P(A | B) = 0.128/0.288

P(A | B) = 0.44 approximately

This is what you have shown in your steps. This means if we know the person is on the honor roll, then they have a 44% chance of getting the class they want.

Those on the honor roll are at a disadvantage to getting their requested class. Perhaps the thinking is that the honor roll students can handle harder or less popular teachers.

Regardless of motivations, being on the honor roll changes the probability of getting the class you want. So Alex is correct in thinking the honor roll students have a disadvantage. Everything would be fair if P(A | B) = P(A) showing that events A and B are independent. That is not the case here so the events are linked somehow.

8 0

3 years ago

Evaluate f(x - 2) when f(x) = 4x² - 4x – 2

ivann1987 [24]

Answer:

Step-by-step explanation:

8 0

2 years ago