All categories

3 years ago

Based on a study, a theme park has reported that typical patrons were willing to pay

Mathematics

1 answer:

Damm [24]3 years ago

3 0

Answer:

A and D

Step-by-step explanation:

You might be interested in

The sides of a square have length s.<br> 4 Write a formula for the perimeter P of the square.

OLga [1]

Answer:P= s(4)

Step-by-step explanation:

8 0

2 years ago

Someone help me with this question please

Naily [24]

Answer: $t\geq 1$

Step-by-step explanation:

you would multiply by 3 on both sides to get the division to go away.

then subtract 8 from both sides.

then you have your answer

5 0

3 years ago

O is the midpoint of segment FG. FO=3x+17, OG=7x-15. <br><br><br> 98 points

AlladinOne [14]

O is the midpoint so both sides FO and OG need to equal each other.

3x + 17 = 7x - 15

Subtract 3x from both sides:

17 = 4x -15

Add 15 to both sides:

32 = 4x

Divide both sides by 4:

x = 32/4

X = 8

Now you have the value of X, solve for the lengths:

FO = 3x +17 = 3(8) +17 = 24 +17 = 41

OG = 7x -15 = 7(8) -15 = 56-15 = 41

Total length would be 41 + 41 = 82

5 0

2 years ago

Read 2 more answers

Which statement best describes the function h(t) = 210 – 15t?

jonny [76]

Answer:

The answer is b.

h is the function name; t is the input and h(t) is the output.

The function name is the one outside the parentheses, the input is inside and the output is the whole function.

Step-by-step explanation:

8 0

3 years ago

Find the max and min values of f(x,y,z)=x+y-z on the sphere x^2+y^2+z^2=81

Anton [14]

Using Lagrange multipliers, we have the Lagrangian

$L(x,y,z,\lambda)=x+y-z+\lambda(x^2+y^2+z^2-81)$

with partial derivatives (set equal to 0)

$L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}$
$L_y=1+2\lambda y=0\implies y=-\dfrac1{2\lambda}$
$L_z=-1+2\lambda z=0\implies z=\dfrac1{2\lambda}$
$L_\lambda=x^2+y^2+z^2-81=0\implies x^2+y^2+z^2=81$

Substituting the first three equations into the fourth allows us to solve for $\lambda$ :

$x^2+y^2+z^2=\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}=81\implies\lambda=\pm\dfrac1{6\sqrt3}$

For each possible value of $\lambda$ , we get two corresponding critical points at $(\mp3\sqrt3,\mp3\sqrt3,\pm3\sqrt3)$ .

At these points, respectively, we get a maximum value of $f(3\sqrt3,3\sqrt3,-3\sqrt3)=9\sqrt3$ and a minimum value of $f(-3\sqrt3,-3\sqrt3,3\sqrt3)=-9\sqrt3$ .

5 0

3 years ago