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2 years ago

Which one of these represents the graph in this picture right here ? A,B,C,D,E or F?

Mathematics

1 answer:

arlik [135]2 years ago

5 0

Answer:

A) y=3x+1

Step-by-step explanation:

Use the formula y=mx+b

m is the slope and b is the y intercept

the graph starts on 1 on the y axis so you replace 1 for b

the slope goes up 3 and over one, giving the fraction 3/1 or just 3. so you replace 3 for m.

You might be interested in

One week the music store sold 2 trumpets, 3 clarinets, and 5 violins for $1,240. The next week, they sold 3 trumpets, 1 clarinet

garik1379 [7]

Answer:

Cost of Trumpets: $174

Cost of Clarinets: $149

Cost of Violins: $89

Step-by-step explanation:

Got it correct

4 0

2 years ago

Easy Cooking is a cooking supply company. Last year, the company sold a total of 5,340 ovens. This year, they sold 7,743 ovens.

jenyasd209 [6]

Answer:

45%

Step-by-step explanation:

Given data

Last year, the company sold a total of 5,340 ovens

This year, they sold 7,743 ovens

Percent increase= final-initial/initial*100

substitute

Percent increase= 7,743-5,340/5,340*100

Percent increase= 2403/5,340*100

Percent increase=0.45*100

Percent increase=45%

Hence the increase is 45%

7 0

2 years ago

Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = yzi + 4xzj + ex

natima [27]

Answer:

The result of the integral is 81π

Step-by-step explanation:

We can use Stoke's Theorem to evaluate the given integral, thus we can write first the theorem:

$\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec S$

Finding the curl of F.

Given $F(x,y,z) = < yz, 4xz, e^{xy} >$ we have:

$curl \vec F =\left|\begin{array}{ccc} \hat i &\hat j&\hat k\\ \cfrac{\partial}{\partial x}& \cfrac{\partial}{\partial y}&\cfrac{\partial}{\partial z}\\yz&4xz&e^{xy}\end{array}\right|$

Working with the determinant we get

$curl \vec F = \left( \cfrac{\partial}{\partial y}e^{xy}-\cfrac{\partial}{\partial z}4xz\right) \hat i -\left(\cfrac{\partial}{\partial x}e^{xy}-\cfrac{\partial}{\partial z}yz \right) \hat j + \left(\cfrac{\partial}{\partial x} 4xz-\cfrac{\partial}{\partial y}yz \right) \hat k$

Working with the partial derivatives

$curl \vec F = \left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(4z-z\right) \hat k\\curl \vec F = \left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(3z\right) \hat k$

Integrating using Stokes' Theorem

Now that we have the curl we can proceed integrating

$\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec S$

$\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot \hat n dS$

where the normal to the circle is just $\hat n= \hat k$ since the normal is perpendicular to it, so we get

$\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S \left(\left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(3z\right) \hat k\right) \cdot \hat k dS$