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qwelly [4]
3 years ago
10

Whoever answers I’ll mark brainliest!

Mathematics
1 answer:
harina [27]3 years ago
8 0

Assuming that there is no dots on any of the bottom shown, i would go with the 2nd top left, that would be the next pattern.

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A salesman receives a weekly salary of $150. He
shusha [124]

Answer:

150 + 4x = 330

4x = 180

x = 45

Step-by-step explanation:

8 0
3 years ago
Am not sure how to solve this problem (number 13)
Andrej [43]

check the picture below.


a positive slope, means an increasing slope, namely, a line that is higher from left-to-right, so the right tip of it is going upwards. A negative y-intercept,


now, if you notice the 3 green lines, the line is increasing, the right-side is higher than the left-side, and the y-intercept, namely where x = 0, or where the graph touches the y-axis, their y-intercept is on a negative value of y.


now, look at the red line, it has also a positive slope, is increasing, however, there's no way it can have negative y-intercept. If it remains increasing, it can only have a y-intercept of 3 or higher, but not lower than that. The only way it can have a negative value for the y-intercept is if it tips over, but if we do that, the slope will be decreasing.

4 0
4 years ago
Calculate the rise and run and find the slope ( (4,-7) and (4,0)
Marta_Voda [28]
Hey there! :D

Slope formula:

m= (y2-y1)/(x2-x1)

m= (0-(-7)/(4-4)

m=0+7/0

m=7/0

The rise is 7 and the run is 0.

I hope this helps!
~kaikers
7 0
4 years ago
[6x(5-3)+9]-5 with step by step
Arada [10]

Answer: 10

Step-by-step explanation: [30-24+9]=15-5=10

7 0
3 years ago
Help me about this integral
NemiM [27]

The gradient theorem applies here, because we can find a scalar function <em>f</em> for which ∇ <em>f</em> (or the gradient of <em>f</em> ) is equal to the underlying vector field:

\nabla f(x,y,z)=\langle2xy,x^2-z^2,-2yz\rangle

We have

\dfrac{\partial f}{\partial x}=2xy\implies f(x,y,z)=x^2y+g(y,z)

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

\dfrac{\partial f}{\partial z}=-2yz=-2yz+\dfrac{\mathrm dh}{\mathrm dz}\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C

where <em>C</em> is an arbitrary constant.

So we found

f(x,y,z)=x^2y-yz^2+C

and by the gradient theorem,

\displaystyle\int_{(0,0,0)}^{(1,2,3)}\nabla f\cdot\langle\mathrm dx,\mathrm dy,\mathrm dz\rangle=f(1,2,3)-f(0,0,0)=\boxed{-16}

5 0
3 years ago
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