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jekas [21]
2 years ago
10

How many points of intersection does this system have? y = 3 x minus 4. Negative 3 x y = 4. Zero points of intersection one poin

t of intersection two points of intersection an infinite number of points of intersection.
Mathematics
1 answer:
defon2 years ago
7 0

A point of intersection is a point where two lines or curves meet.

There are zero points of intersection.

Given

The equations are;

\rm y = 3x-4\\\\-3x+y=4

<h3>What is the point of intersection?</h3>

A point of intersection is a point where two lines or curves meet.

There are two methods to find a point of intersection graphically by graphing the curves on the same graph and identifying their points and another one is to solve the equation and find the value of x and y.

Adding both the equation

\rm y-3x-3x+y=-4+4\\\\2y=6x\\\\y=3x

Substitute the value of y in the equation

\rm y=3x-4\\\\3x=3x-4\\\\3x-3x=-4\\\\0=-4

Hence, there are zero points of intersection.

To know more about the point of intersection click the link given below.

brainly.com/question/7039469

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Let f(x)= 3x2 +5 and g(x) = - 5x + 3. Find the following.<br> (f-g)(-4)<br> (f-9)(-4)=
masha68 [24]
Answer: 3x^2 - 3

Step by step explanation:
5 0
3 years ago
A cylinder shaped can needs to be constructed to hold 500 cubic centimeters of soup. The material for the sides of the can costs
iogann1982 [59]

Answer:

r=3.628cm

h=12.093cm

Step-by-step explanation:

For this problem we are going to use principles, concepts and calculations from multivariable calculus; mainly we are going to use the Lagrange multipliers method. This method is thought to help us to find a extreme value of a multivariable function 'F' given a restriction 'G'. F represents the function that we want to optimize and G is just a relation between the variables of which F depends. The Lagrange method for just one restriction is:

\nabla F=\lambda \nabla G

First, let's build the function that we want to optimize, that is the cost. The cost is a function that must sum the cost of the sides material and the cost of the top and bottom material. The cost of the sides material is the unitary cost (0.03) multiplied by the sides area, which is A_s=2\pi rh for a cylinder; while the cost of the top and bottom material is the unitary cost (0.05) multiplied by the area of this faces, which is A_{TyB}=2\pi r^2 for a cylinder.

So, the cost function 'C' is:

C=2\pi rh*0.03+2\pi r^2*0.05\\C=0.06\pi rh+0.1\pi r^2

The restriction is the volume, which has to be of 500 cubic centimeters:

V=500=\pi r^2h\\500=\pi hr^2

So, let's apply the Lagrange multiplier method:

\nabla C=\lambda \nabla V\\\frac{\partial C}{\partial r}=0.06\pi h+0.2\pi r\\\frac{\partial C}{\partial h}=0.06\pi r\\\frac{\partial V}{\partial r}=2\pi rh\\\frac{\partial V}{\partial h}=\pi r^2\\(0.06\pi h+0.2\pi r,0.06\pi r)=\lambda (2\pi rh,\pi r^2)

At this point we have a three variable (h,r, λ)-three equation system, which solution will be the optimum point for the cost (the minimum). Let's write the system:

0.06\pi h+0.2\pi r=2\lambda \pi rh\\0.06\pi r=\lambda \pi r^2\\500=\pi hr^2

(In this kind of problems always the additional equation is the restricion, in this case, V=500).

Let's divide the first and second equations by π:

0.06h+0.2r=2\lambda rh\\0.06r=\lambda r^2\\500=\pi hr^2

Isolate λ from the second equation:

\lambda =\frac{0.06}{r}

Isolate h from the third equation:

h=\frac{500}{\pi r^2}

And then, replace λ and h in the first equation:

0.06*\frac{500}{\pi r^2} +0.2r=2*(\frac{0.06}{r})r\frac{500}{\pi r^2} \\\frac{30}{\pi r^2}+0.2r= \frac{60}{\pi r^2}

Multiply all the resultant equation by \pi r^{2}:

30+0.2\pi r^3=60\\0.2\pi r^3=30\\r^3=\frac{30}{0.2\pi } =\frac{150}{\pi}\\r=\sqrt[3]{\frac{150}{\pi}}\approx 3.628cm

Then, find h by the equation h=\frac{500}{\pi r^2} founded above:

h=\frac{500}{\pi r^2}\\h=\frac{500}{\pi (3.628)^2}=12.093cm

4 0
3 years ago
A car travels 370 1/2 miles in 6 1/2 hours what is the average speed at which the car is traveling
Alex

Answer:

57 miles per hour

Step-by-step explanation:

Speed is equal to distance over time

I changed the fractions to decimals for ease of math

distance = 370 .5 miles

time = 6.5 hours

Speed = 370.5/6.5 = 57 miles/hour

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4 years ago
Select two ratios that are equivalent to
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Answer:

21 : 30 and 35 : 50​

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X=<br> A) 80<br> B) 90<br> C) 100
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X= A) 80 I believe you double 120 to 240 and divide by the 4 quarters.
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4 years ago
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