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romanna [79]
3 years ago
9

Which function below is the inverse of f(x) = x2 − 25?

Mathematics
2 answers:
FrozenT [24]3 years ago
8 0
(x-5)(x+5) is the answer
IRINA_888 [86]3 years ago
3 0
f(x)  =   x^{2}  -  25
  can be written as y =  x^{2}  -  25

when finding inverse the first step is to interchange x and y in the equation

∴ (x) = (y)^{2}  -  25

then you simply solve for y

x  +   25  =   y^{2}

y  =   \sqrt{ (x + 25)}

Thus f^{-1} (x)  =   \sqrt{(x  +  25)}
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Find an equation of the tangent to the curve x =5+lnt, y=t2+5 at the point (5,6) by both eliminating the parameter and without e
svet-max [94.6K]

ANSWER

y = 2x -4

EXPLANATION

Part a)

Eliminating the parameter:

The parametric equation is

x = 5 +  ln(t)

y =  {t}^{2}  + 5

From the first equation we make t the subject to get;

x - 5 =  ln(t)

t =  {e}^{x - 5}

We put it into the second equation.

y =  { ({e}^{x - 5}) }^{2}  + 5

y =  { ({e}^{2(x - 5)}) }  + 5

We differentiate to get;

\frac{dy}{dx}  = 2 {e}^{2(x - 5)}

At x=5,

\frac{dy}{dx}  = 2 {e}^{2(5 - 5)}

\frac{dy}{dx}  = 2 {e}^{0}  = 2

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

y-y_1=m(x-x_1)

y - 6 = 2(x - 5)

y = 2x - 10 + 6

y = 2x -4

Without eliminating the parameter,

\frac{dy}{dx}  =  \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }

\frac{dy}{dx}  =  \frac{ 2t}{  \frac{1}{t} }

\frac{dy}{dx}  =  2 {t}^{2}

At x=5,

5 = 5 +  ln(t)

ln(t)  = 0

t =  {e}^{0}  = 1

This implies that,

\frac{dy}{dx}  =  2 {(1)}^{2}  = 2

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

y-y_1=m(x-x_1)

y - 6 = 2(x - 5) =

y = 2x -4

5 0
3 years ago
Given that a triangle has two sides of lengths 5 inches and 7 inches, which is a possible length of the third side?
victus00 [196]

Answer:

big poop face looking butt with a massive forehead poopy duck dipers

Step-by-step explanation:

↑∴↔↔⇔⇔³∛÷≈≤≥≥β⇵⊕⊕¬⊕⊕⊕⊕⊕⊕¬∪∪⊥∡∠°∞㏒㏑∡Ф⇒⇒√³π∫↓∵∴∴\lim_{n \to \infty} a_n  \lim_{n \to \infty} a_n \neq \sqrt{x} x^{2} x^{2} \geq \geq \pi \frac{x}{y} and thats why it is big poop face looking butt with a massive forehead poopy duck dipers

3 0
3 years ago
The domain of u(x) is the set of all real values except 0 and the domain of v(x) is the set of all real values except 2. What
Oksana_A [137]

Using the domain concept, the restrictions on the domain of (u.v)(x) are given by:

A. u(x) ≠ 0 and v(x) ≠ 2.

<h3>What is the domain of a data-set?</h3>

The domain of a data-set is the set that contains all possible input values for the data-set.

To calculate u(x) x v(x) = (u.v)(x), we calculate the values of u and v and then multiply them, hence the restrictions for each have to be considered, which means that statement A is correct.

Summarizing, u cannot be calculated at x = 0, v cannot be calculated at x = 2, hence uv cannot be calculated for either x = 0 and x = 2.

More can be learned about the domain of a data-set at brainly.com/question/24374080

#SPJ1

3 0
2 years ago
If a fire is burning a building and it takes 2 hours to put it out using 2000 gal./min., what minimum diameter cylindrical tank
strojnjashka [21]

Answer:

The minimum diameter of the cylindrical tank needed to store the quantity needed to put out the fire is approximately 58.415 feet.

Step-by-step explanation:

A gallon equals 0.134 cubic feet. First, we determine the amount of water (Q), measured in cubic feet, needed to put out the fire under the assumption that water is consumed at constant rate:

Q = \dot Q \cdot \Delta t (1)

Where:

\dot Q - Volume rate, measured in feet per minute.

\Delta t - Time, measured in minutes.

If we know that \dot Q = 2000\,\frac{gal}{min} and \Delta t = 120\,min, then the amount of water is:

Q = \left(2000\,\frac{gal}{min} \right)\cdot (120\,min) \cdot \left(0.134\,\frac{ft^{3}}{gal} \right)

Q = 32160\,ft^{3}

And the diameter of the cylindrical tank based on the capacity found above is determined by volume formula for a cylinder:

Q = \frac{\pi}{4}\cdot D^{2}\cdot h (2)

Where:

D - Diameter, measured in feet.

h - Height, measured in feet.

If we know that Q = 32160\,ft^{3} and h = 12\,ft, then the minimum diameter is:

D^{2} = \frac{4\cdot Q}{\pi\cdot h}

D = 2\cdot \sqrt{\frac{Q}{\pi\cdot h} }

D = 2\cdot \sqrt{\frac{32160\,ft^{3}}{\pi\cdot (12\,ft)} }

D \approx 58.415\,ft

The minimum diameter of the cylindrical tank needed to store the quantity needed to put out the fire is approximately 58.415 feet.

8 0
3 years ago
Simplify <br>3x2 + x + 2x2​
jenyasd209 [6]

Answer:

10+ x

Step-by-step explanation:

Follow me please for answering the question

8 0
3 years ago
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