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3 years ago

F(x)=2(x)^2+5 (x+2) f(2)

Mathematics

1 answer:

ehidna [41]3 years ago

3 0

Answer:

−2x^2+y /10x+20

Step-by-step explanation:

Let's solve for f.

y=2x^2+5(x+2)f(2)

Step 1: Flip the equation.

10fx+2x^2+20f=y

Step 2: Add -2x^2 to both sides.

10fx+2x^2+20f+−2x^2=y+−2x^2

10fx+20f=−2x^2+y

Step 3: Factor out variable f.

f(10x+20)=−2x^2+y

Step 4: Divide both sides by 10x+20.

f(10x+20) /

10x+20

−2x2+y

10x+20

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Ben works for software company in Cleveland. To train new employees, Ben travels to their two other locations, Buffalo and Phila

Andru [333]

Answer:

Explanation:we will solve this question with help of linear programming.

Now, let Ben goes x times in buffalo trip and goes y times in Philadelphia trip. Thus, according to the question, Min Z= 180x+300y

And s.t.r. 2x+3y≤105, where x≥5, y≥15.

since at (0,0) 0≤105(true), so area of 2x+3y≤105 will cover the origin.

But, 0≥5 and 0≥15 are false, so their area will not cover the origin.

Thus, after drawing the graph of these three lines we will get a feasible reason of triangle ABC, where point A≡(5,31.667), B≡(5,15), and C≡(30,15)

Z at A- 180×5+300×31.667=10,400.1

at B-180×5+300×15=1350( which is minimum.)

at C- 180×30+300×15=5,850

Thus, Ben should go 5 times to buffalo trip and 15 times to Philadelphia to minimize his cost which is $1350.

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3 years ago

Two particles are fixed to an x axis: particle 1 of charge -1.00 x 10-7 C is at the origin and particle 2 of charge +1.00 x 10-7

r-ruslan [8.4K]

The electric field strength at any point from a charged particle is given by E = kq/r^2 and we can use this to calculate the field strength of the two fields individually at the midpoint.

The field strength at midway (r = 0.171/2 = 0.0885 m) for particle 1 is E = (8.99x10^9)(-1* 10^-7)/(0.0885)^2 = -7.041 N/C and the field strength at midway for particle 2 is E = (8.99x10^9)(5.98* 10^-7)/(0.0935)^2 = <span>-7.041 N/C
</span>
Note the sign of the field for particle 1 is negative so this is attractive for a test charge whereas for particle 2 it is positive therefore their equal magnitudes will add to give the magnitude of the net field, 2*<span>7.041 N/C </span>= 14.082 N/C

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3 years ago

A family on vacation records the time in hours) and the distance (in miles) that they travel. The data are shown in the scatter

nata0808 [166]

Answer:

In the first 3.5 hours, the family travelled 175 miles.

Step-by-step explanation:

We have the table:

time (hours) distance ( miles)

0 0

1 50

2 100

3 150

4 200

5 250

6 300

7 350

8 400

9 450

This seems like a linear relationship, let's write the general line:

y = a*x + b

That represents this situation, y is the distance in miles and x is the time in hours.

In the line equation, a represents the slope and b the y-intercept.

And if the line passes through the points (x₁, y₁) and (x₂, y₂), then the slope of the line is:

a = (y₂ - y₁)/(x₂ - x₁)

In this case we have a lot of points for the line, we can use any pair we want (we actually should use a few different pairs to see that the slope is always the same)

So if we use the first two (0, 0) and (1, 50)

the slope is:

a = (50 - 0)/(1 - 0) = 50

Now, if we use the third and fifth points (4, 200) and (5, 250) now the slope is:

a = (250 - 200)/(5 - 4) = 50/1 = 50

So we can conclude that the slope of the equation is a = 50

Then the equation is something like:

y = 50*x + b

To find the value of b, we can use any of the points of the line, for example (0, 0)

This means that when x = 0, y also must be 0, then:

0 = 50*0 + b

0 = b

Then the equation is:

y = 50*x

Now, to answer how many miles the family travelled in the first 3.5 hours, we just need to replace x by 3.5 in the equation:

y = 50*3.5 = 175

In the first 3.5 hours, the family travelled 175 miles.

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In ADEF, the measure of

MrMuchimi

Answer:

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You can use a protractor

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TiliK225 [7]

I know of two ways to solve quadratic equations. The first is through factoring. Let us take the example (x^2)+2x+1=0. We can factor this equation out and the factors would be (x+1)(x+1)=0. To solve for the roots, we equate each factor to 0, that is

x+1=0; x+1=0

In this case, the factors are the same so the root of the equation is

x=1.

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3 years ago