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fgiga [73]
3 years ago
9

Plss helpppppppppp w,jvkwsvkjbw,vj effvdefv

Mathematics
1 answer:
JulsSmile [24]3 years ago
4 0

Answer:180

Step-by-step explanation:

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In 1985 there were 275 cell phone subscribers in the small town of China grove. Subscribers increased by 40% each year after 198
emmasim [6.3K]
1985year ->275cell phones->100%
1994year -> x cell phones ->100%+360%
40% • (1994-1985)= 40% • 9= 360%

275 cell phone 100%
X cell phones. 100+360%=460%
x=275•460%/100%
x=12420/100
x=1265
99% sure this is a correct answer


x=275 • 460%/100%
x=12,420/100
x=1265

1994-1985= 9 yrs • 40%= 360%

5 0
2 years ago
Nth term (math) asap please.
SOVA2 [1]

Answer:

Just do minus two, and you will get 27-100=-73

Step-by-step explanation:

7 0
2 years ago
In the coordinate plane, choose the graph with the conditions given. x + y = 10
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Could you post a picture with the graphs in it? Kinda hard to answer your question without the answer choices :)
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3 years ago
Read 2 more answers
Let S be the surface defined by x 2 + 2y 3 + 3z 4 = 6. Let T be the surface defined parametrically by r(u, v) = (1+ln u, 2e v+u−
aleksandrvk [35]

The tangent to C through (1, 1, 1) must be perpendicular to the normal vectors to the surfaces S and T at that point.

Let f(x,y,z)=x^2+2y^3+3z^4. Then S is the level curve f(x,y,z)=6. Recall that the gradient vector is perpendicular to level curves; we have

\nabla f(x,y,z)=(2x,6y,12z^2)

so that the gradient of f at (1, 1, 1) is

\nabla f(1,1,1)=(2,6,12)

For the surface T, we have

\begin{cases}1+\ln u=1\\2e^v+u-2=1\\uv+1=1\end{cases}\implies u=1,v=0

so that \vec r(1,0)=(1,1,1). We can obtain a vector normal to T by taking the cross product of the partial derivatives of \vec r(u,v), and evaluating that product for u=1,v=0:

\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left(u-2ve^v,-1,\dfrac{2e^v}u\right)

\left(\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right)(1,0)=(1,-1,2)

Now take the cross product of the two normal vectors to S and T:

(2,6,12)\times(1,-1,2)=(24,8,-8)

The direction of vector (24, 8, -8) is the direction of the tangent line to C at (1, 1, 1). We can capture all points on the line containing this vector by scaling it by t\in\Bbb R. Then adding (1, 1, 1) shifts this line to the point of tangency on C. So the tangent line has equation

\vec\ell(t)=(1,1,1)+t(24,8,-8)=(1+24t,1+8t,1-8t)

7 0
3 years ago
What is another name for plane Z?
lara31 [8.8K]
Other names for "plane Z" include:
___________________________
1) The "z-plane"; 
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2) The "complex plane";
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3) <span>the "domain of the </span>z<span>-transform"
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2 years ago
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