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2 years ago

if f(x) and it’s inverse function, f-1(x) are both plotted on the same coordinate plane, what is their point of intersection?

Mathematics

2 answers:

Zanzabum2 years ago

6 0

Answer: The correct answer is D) (3,3) took the quiz and got it right.

Alenkasestr [34]2 years ago

4 0

For this case we observe first the axis of symmetry of both functions.
We note that the axis is:
x = 1
Then, the point of intersection of both graphs is the following ordered pair:
(x, y) = (1, -1)
Answer:
(x, y) = (1, -1)

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Solve this problem please

djyliett [7]

The formula for depreciation is:

$y=x(1-r)^t$

Where x = Initial value,

y= Amount after depreciation.

r= Rate of depreciation,

t = time (in years)

According to given problem,

x = 1040, y= 944 and t = 12 months =1 year.

So, first step is to plug in these values in the above formula, So,

$944 = 1040(1-r)^1$

944 = 1040 (1 -r)

$\frac{944}{1040} =1-r$ Divide each sides by 1040.

0.907692308 =1 - r

0.907692308 - 1 = -r Subtract 1 from each sides.

-0.092307692 = -r

So, r = 0.09 or 9%.

Now plug in 0.09 in the above equation to get the depreciation equation. So,

$y = x (1- 0.09)^t$

So, $y = 0.91x^t$

b) To find the value of the bike after 5 months,

plug in t = 5 months= 5/12 = 0.41667 years in the above equation of depreciation.

So, $y= 1040 *(0.91)^{0.41667}$

y = 1040 * 0.961465659

y = 999.9242852

y = 1000 (Rounded to nearest integer).

Hence, the value of the bike after 5 months is $1000.

7 0

3 years ago

Find the equations of the tangent lines to x2+y2=1 at x=0, x=1 and x=35.

Ivanshal [37]

Assuming the equation is x^2 + y^2 = 1, then that's a circle, with radius 1, centered on the origin [0,0].

So there are two tangents at x = 0. They are y = 1, and y = -1 (horizontal lines).

There is one tangent at x = 1. It is x = 1 (a vertical line).

There is no tangent at x = 35, because the original equation has no solution at x = 35.

4 0

3 years ago

Dillon says that 4b and -2b are not like terms because 4b is positive and -2b is negative. Is he correct? Explain.

Novosadov [1.4K]

No, that is not a factor to make it not a like term.

6 0

3 years ago

Read 2 more answers

Are the points U, H, and L collinear? U E S H

butalik [34]

Answer:

Yes