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3 years ago

At which values of X does the function F(x) have a vertical asymptote check all apply F(x) =3/(x+2) (x-7)

Mathematics

1 answer:

finlep [7]3 years ago

8 0

Answer:

When x = -2 or when x = 7

Step-by-step explanation:

x+2 is a factor in the denominator. If x = -2, the factor becomes zero.

x-7 is a factor in the denominator. If x = 7, the factor becomes zero.

We can't have zero as a factor in a denominator because division by zero is not defined. It doesn't have any meaning.

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What is the correct answer to this question?<br><br> F/2=5/8

ANTONII [103]

Answer:

F = 1.25

Step-by-step explanation:

All you would have to do is cross multiply, so our equations starts at $\frac{F}{2} = \frac{5}{8}$ . We would now multiply F by 8 and 2 by 5. We now have 8F = 10. We can now divide both sides of the equation by 8 to get F = 1.25.

4 0

3 years ago

Please help me with this

MissTica

Answer:

For the question on the right, I assume b is the exponential factor, or growth factor.

Step-by-step explanation:

Linear:

$Y=aX + b\\(X, Y) = (0, 15) => b = 15\\(X,Y) = (1, 12) => a = -3\\Y = -3X + 15$

You can find the equation by using two points and the equation for a line.

Exponential:

$Y = a^{bX+c}\\(X, Y) = (0, 81) => a^{c} = 81 => a^{c} = 3^{4} => a=3, c=4\\(X, Y) = (1, 27) => a^{b+c} = 27 => a^{b}a^{c} = 27 => a^{b} = \frac{27}{81} = \frac{1}{3} => b=-1\\Y = 3^{-X+4}$

6 0

3 years ago

Show that the line integral is independent of path by finding a function f such that ?f = f. c 2xe?ydx (2y ? x2e?ydy, c is any p

Juli2301 [7.4K]

I'm reading this as

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy$

with $\nabla f=(2xe^{-y},2y-x^2e^{-y})$ .

The value of the integral will be independent of the path if we can find a function $f(x,y)$ that satisfies the gradient equation above.

You have

$\begin{cases}\dfrac{\partial f}{\partial x}=2xe^{-y}\\\\\dfrac{\partial f}{\partial y}=2y-x^2e^{-y}\end{cases}$

Integrate $\dfrac{\partial f}{\partial x}$ with respect to $x$ . You get

$\displaystyle\int\dfrac{\partial f}{\partial x}\,\mathrm dx=\int2xe^{-y}\,\mathrm dx$
$f=x^2e^{-y}+g(y)$

Differentiate with respect to $y$ . You get

$\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]$
$2y-x^2e^{-y}=-x^2e^{-y}+g'(y)$
$2y=g'(y)$

Integrate both sides with respect to $y$ to arrive at

$\displaystyle\int2y\,\mathrm dy=\int g'(y)\,\mathrm dy$
$y^2=g(y)+C$
$g(y)=y^2+C$

So you have

$f(x,y)=x^2e^{-y}+y^2+C$

The gradient is continuous for all $x,y$ , so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy=f(4,1)-f(1,0)=\frac9e$

8 0

3 years ago

12.5 = 3x = 20

kotegsom [21]

The correct answer would be B

7 0

3 years ago

When flying at an altitude of 5 miles, the lines of sight to the horizon looking north and south make about a 173.7 degree angle

oksian1 [2.3K]

5 miles high is one of the sides of a triangle depending on accuracy level
h^2=x^2+y^2
we don't have 2 distances
Tan A=O/a
O=a tan A
We solve for O because the angle is at the top of the line going up and we want the opposite angle that is along the ground
O=5×tan(173.7/2)=90.854033512
The distance he can see is:
90.85*2~181.7 miles

Now we need to find the distance between lines:
The north south distance between each line is 69 miles
thus the number of degrees he will see will be:
181.7/69
=2 19/30

7 0

3 years ago