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densk [106]
2 years ago
6

10

Mathematics
1 answer:
Natasha2012 [34]2 years ago
6 0

Answer:

x= -2

Step-by-step explanation:

Rational Function can be expressed as:

\displaystyle \large{y =  \frac{a}{p(x - b)}  + q}

Where b is our horizontal shift and q is our vertical shift.

For asymptotes, there are two types which are vertical and horizontal.

Vertical Asymtote is the value of -b itself or x = -b.

Therefore, from the functio:

\displaystyle   \large{y =  \frac{10}{x + 2} }

Our vertical asymtote is x = -2

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Is the given ordered pair a solution to the equation? <br><br><br> y=2x+7;(-2, 3)
iren2701 [21]

Answer:

The answer to your question is: Yes, it is a solution

Step-by-step explanation:

           

Point  (-2, 3)

                         Line:            y = 2x + 7

Process, replace the point in the line

                                          3 = 2(-2) + 7

                                          3 = -4 + 7

                                          3 = 3

As we got that 3 equals 3, then the point given is a solution of the equation.

               

5 0
3 years ago
1. If f(x) = 4x + 5, what is the value of f(-3)? *
Dmitry [639]
-18 hope this doesn’t get hate from people that a gods at this app and don’t know how to mind there beeswax
5 0
2 years ago
Read 2 more answers
The diagram shows a parallelogram.
Firlakuza [10]

Check the picture below.

4 0
2 years ago
Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i
sp2606 [1]

i. The Lagrangian is

L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)

with critical points whenever

L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4

L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1

L_\lambda=4x^2+y^2-9=0

  • If x=0, then L_\lambda=0\implies y=\pm3.
  • If y=0, then L_\lambda=0\implies x=\pm\dfrac32.
  • Either value of \lambda found above requires that either x=0 or y=0, so we get the same critical points as in the previous two cases.

We have f(0,-3)=9, f(0,3)=9, f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25, and f\left(\dfrac32,0\right)=\dfrac{729}4, so f has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)

with critical points whenever

L_x=2\lambda x=0\implies x=0 (because we assume \lambda\neq0)

L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1

L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda

L_\lambda=x^2+y^2+z^2-36=0

  • If x=y=0, then L_\lambda=0\implies z=\pm6.
  • If \lambda=-1, then z=-5, and with x=0 we have L_\lambda=0\implies y=\pm\sqrt{11}.

We have f(0,0,-6)=60, f(0,0,6)=-60, f(0,-\sqrt{11},-5)=61, and f(0,\sqrt{11},-5)=61. So f has a maximum value of 61 and a minimum value of -60.

5 0
3 years ago
Lisa walks at a speed of 3 25 miles per hour for 1 5 hours How many miles does she walk in all?
Blababa [14]

Answer:

I think it might be 21.6 or just 21

7 0
2 years ago
Read 2 more answers
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