All categories

4 years ago

There are 19763 tickets

Mathematics

1 answer:

emmainna [20.7K]4 years ago

6 0

So what yeah there are 19763 tickets

You might be interested in

The product of 14 and a number is 140

BigorU [14]

Hey!

------------------------------------------------

Solution #1:

Write an expression.

14 × y = 140

Simplify.

14y = 140

Divide both sides by 14.

14y/14 = 140/14

Answer.

y = 10

------------------------------------------------

Solution #2:

Divide.

140 / 14

Answer.

------------------------------------------------

Answer:

The product of 14 and 10 is 140.

------------------------------------------------

Hope This Helped! Good Luck!

6 0

4 years ago

For #36, why is the answer 4? Plz explain.

luda_lava [24]

I am getting 7 as the value of b. I do not believe 4 is a correct answer!

Here's why:

General equation of a circle in the xy plane

$(x-c_x)^2+(y-x_y)^2=r^2$

with $c_x\,\,\mbox{and}\,\,c_y$ being the x,y coordinates of the center.

The center point can be determined from the two endpoints of the diameter because the line segment is horizontal, so we just find the center of the x coordinates of the two points:

$c_x = -2+\frac{|8-(-2)|}{2}=3\\c_y = 3$

and the radius r = diameter/2 = 8-(-2)/2 = 5

so we have

$(x-3)^2+(y-3)^2=5^2$

Given a point on the circle (0,b) with b>0:

$(0-3)^2+(b-3)^2=5^2\\9+(b-3)^2 = 25\\(b-3)^2 = 16\\b=7$

6 0

3 years ago

Find the extreme values of the function f(x, y) = 4x2 + 6y2 on the circle x2 + y2 = 1.

julia-pushkina [17]

Answer with Step-by-step explanation:

We are given that

$f(x,y)=4x^2+6y^2$

Let g(x,y)= $x^2+y^2=1$

We have to find the extreme values of the given function

$\nabla f(x,y)$

$\nabla g(x,y)=$

Using Lagrange multipliers

$\nabla f(x,y)=\lambda \nabla g(x,y)$

$f_x=\lambda g_x$

$8x=\lambda 2x$

Possible value x=0 or $\lambda=4$

If x=0 then substitute the value in g(x,y)

Then, we get $y=\pm 1$

$f_y=\lambda g_y$

$12y=\lambda 2y$

If $\lambda=4$ and substitute in the equation

Then , we get possible value of y=0

When y=0 substitute in g(x,y) then we get

$x=\pm 1$

Hence, function has possible extreme values at points (0,1),(0,-1), (1,0) and (-1,0).

$f(0,1)=6$

$f(0,-1)=6$

$f(1,0)=4$

$f(-1,0)=4$

Therefore, the maximum value of f on the circle $x^2+y^2=1$ is $f(0,\pm1)=6$ and minimum value of $f(\pm1,0)=4$

7 0

3 years ago

Find the midpoint of a line segment with endpoints (4, -10) and (-8, 2).

pychu [463]

Step-by-step explanation: (-2,-4)

5 0

3 years ago

Solve for x in the diagram below.

enyata [817]

Answer:

Step-by-step explanation:

2x+100=180

-100 -100

2x =80

/2 /2

x =40

3 0

3 years ago

Read 2 more answers