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3 years ago

Twellve concert tickets cost $450 what is the cost of 3 tickets to the same concert

Mathematics

1 answer:

Scrat [10]3 years ago

4 0

1) $450/12=$37.50
2)$37.50x3=$112.50

The answer is: $112.50

I hope this helps

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Determine whether each function is linear or quadratic. Identify the quadratic, linear, and constant terms. g(x)=-2 x²-3(x-2)

valentinak56 [21]

The function g(x) is a quadratic function:

Its quadratic term is -2x²
Its linear term is - 3x
Its constant is 6

<h3>What are functions?</h3>

Functions are algebraic expressions that have at least two variables, in order to make their visual representation through a graph and evaluate their behavior.

This problem deals with linear or quadratic functions, and these differ in the degree of the exponent of the variable:

1 : linear

2 : quadratic

The function:

g(x) = -2x² - 3(x- 2).

g(x) = -2x² - 3x + 6 , we have a quadratic function.

Its quadratic term is -2x²
Its linear term is - 3x
Its constant is 6

Learn more about functions at:

brainly.com/question/28586957

#SPJ4

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1 year ago

0.03, 0.031, 0.1, 0.013 smallest to largest

vitfil [10]

Answer:

0.013

0.031

0.03

0.1

I hope my answer's correct.

3 0

3 years ago

Can someone help me on problem number 8?

kramer

List the coordinates,
Coordinates are shown as (x,y)
If 2+2+7+7 is 16
You make those your side lengths
Coordinates are (9,7) (2,7) (2,5) (9,5)
they maybe slightly off because of the broken peice of paper

8 0

3 years ago

For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f

Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0$

We want to find $f$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)$

$\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C$

$\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)$

$\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C$

$\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0$

so $\vec F$ is not conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0$