Evaluate the expression when m=-6.3 and n= 5.6.<br> -3n+2m

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

Find the value of the greater root of x2 - 6x + 5 = 0 A) -5 B)-1 C)1 D)5

VARVARA [1.3K]

2x-6x+5=0
-4x+5=0
-4x=-5

4 0

3 years ago

Read 2 more answers

Use the distributive property to write 3y - y as a product. Then simplify.

kvasek [131]

Y(3-1)= 2y is the answer to the question

4 0

3 years ago

PLEAASE HELP I NEED UNGROUNDED

omeli [17]

Answer:

y=3x-2

Step-by-step explanation:

The formula for slope is y=mx=b

You can find the slope by finding two points that lie on the line (1,1) & (0,-2)

The formula for slope is $m=\frac{y_{2} -y_{1}}{ x_{2} -x_{1} }$

You plug the values in and you get $m=\frac{1-(-2)}{ 1 -0 }$

Simplify and you get 3

The slope is 3

The y-intercept (b) is -2

4 0

2 years ago

Read 2 more answers

Identify the horizontal asymptote of f(x) = quantity 2 x minus 1 over quantity x squared minus 7 x plus 3.

Rina8888 [55]

We must recall that a horizontal asymptote is the value/s of y that the given function approaches to but never reaches. To find this in a rational function, we compare the expressions with highest degree in the numerator and denominator. There are three possible outcome when this happens.

1. if the highest degree (highest exponent) in the numerator is bigger than that of the denominator, then there won't be any horizontal asymptote.

2. if the highest degree in the denominator is bigger, then the horizontal symptote would be y = 0.

3. if they have the same highest degree, then we just get the quotient of their coefficient.

Now, going back to our function, we have

$f(x) = \frac{2x - 1}{x^{2} - 7x + 3}$

From this we can see that the highest degree in the numerator is 1 (from 2x) and 2 (from x²) for the denominator. Clearly, it shows that its denominator has a higher degree. And from our discussion, we can conclude that the horizontal asymptote would be y = 0.

Answer: y = 0

3 0

3 years ago

Evaluate the expression when m=-6.3 and n= 5.6. -3n+2m