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Masteriza [31]
4 years ago
9

What's the answer? pls help

Mathematics
2 answers:
pishuonlain [190]4 years ago
8 0

Answer:

The slope is 0

Step-by-step explanation:

The slope is the change of y over the change in x. Since y doesn't change at all, it's 0 over 3. 0/3 is just 0, so the slope is 0.

MatroZZZ [7]4 years ago
8 0

Answer:

undefined

Step-by-step explanation:

slope = y2 - y1 / x2 - x1

slope = 5 - 5 / 6 - 3

slope = 0/3

slope = undefined

Undefined because all y coordinates are the same number: 5

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VashaNatasha [74]

Answer:$1000

Step-by-step explanation:

3 0
3 years ago
Whats the formula for the surface area of a triangle?
Alecsey [184]

Answer:

use the formula A = 1/2bh, where A = area, b = base, and h = height. Once you have the areas of all sides and faces, you simply add them together to get the surface area.

Hope this helps!

8 0
3 years ago
Read 2 more answers
A 2-column table with 7 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2, 3. The sec
DochEvi [55]

The statements that are true about the intervals of the continuous function are Options 2, 4  and 5

  • f(x) ≤ 0 over the interval [0, 2].
  • f(x) > 0 over the interval (–2, 0).
  • f(x) ≥ 0 over the interval [2, ).

<h3>What is the statement about?</h3>

Looking at the values given, the intervals which satisfies the condition are known to be:

f(x)<=0 over the interval [0,2]

f(x)>0 over the interval (-2,0)

f(x)>=0 over the interval [2,∞)

Because:

Since the table with x  and f(x) values, we have to examine analyze the table and see the each option that is in line with f(x) or not .

Examine the values of x that is from -3 to 3, the f(x) values are both positive and negative . hence f(x)>0 is false over the interval (-∞,3)

Looking at the the interval from 0 to 2, the f(x) values are 0 and negative. Hence, f(x)<=0 over the interval [0,2]

When you look over the interval (-1,1), the f(x) values are said to be both positive and negative and as such, f(x)<0 is false over the interval (-1,1)

When you look at the interval (-2,0) , the f(x)  is positive and as such, f(x)>0 over the interval (-2,0)

Looking at the interval  [2,∞), f(x) is positive and as such, f(x)>=0 over the interval [2,∞)

Therefore, Option 2, 4 and 5 are correct.

Learn more about interval from

brainly.com/question/14454639

#SPJ1

6 0
2 years ago
A business consultant earns a flat fee for his work as well as an hourly fee. He charges his clients at a rate of $75 per hour.
gregori [183]
The answer would be D. y=75x+300 because if you multiply 75 and 20 you would get 1500 and subtract 1800 for how much was paid and 1500 for the result of multiplying 75 and 20
Answer is D. y=75x+300
5 0
4 years ago
Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)
babunello [35]

\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k

We want to find f(x,y,z) such that \nabla f=\vec F. This means

\dfrac{\partial f}{\partial x}=x^2+y

\dfrac{\partial f}{\partial y}=y^2+x

\dfrac{\partial f}{\partial z}=ze^z

Integrating both sides of the latter equation with respect to z tells us

f(x,y,z)=e^z(z-1)+g(x,y)

and differentiating with respect to x gives

x^2+y=\dfrac{\partial g}{\partial x}

Integrating both sides with respect to x gives

g(x,y)=\dfrac{x^3}3+xy+h(y)

Then

f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)

and differentiating both sides with respect to y gives

y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C

So the scalar potential function is

\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}

By the fundamental theorem of calculus, the work done by \vec F along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it L) in part (a) is

\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}

and \vec F does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them L_1 and L_2) of the given path. Using the fundamental theorem makes this trivial:

\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3

\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4

8 0
3 years ago
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