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vovangra [49]
3 years ago
11

I need help oun this ASAP.

Mathematics
1 answer:
romanna [79]3 years ago
3 0
6.4\cdot10^5-5.4\cdot10^4=
10^5(6.4-0.54)=
10^5\cdot5.86=5.86\cdot10^5
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Find the tangent plane to the given surface of f(x,y)=6- 6/5 x-y at the point (5, -1, 1). Make sure tat your final answer for th
HACTEHA [7]

Answer:

Required equation of tangent plane is z=\frac{6}{5}(x-5y-11).

Step-by-step explanation:

Given surface function is,

f(x,y)=6-\frac{6}{5}(x-y)

To find tangent plane at the point (5,-1,1).

We know equation of tangent plane at the point $(x_0,y_0,z_0)[/tex] is,

z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)\hfill (1)

So that,

f(x_0,y_0)=6-\frac{6}{5}(5+1)=-\frac{6}{5}

f_x=-\frac{6}{5}y\implies f_x(5,-1,1)=\frac{6}{5}

f_y=-\frac{6}{5}x\implies f_y(5,-1,1)=-6

Substitute all these values in (1) we get,

z=\frac{6}{5}(x-5)-6(y+1)-\frac{6}{5}

\therefore z=\frac{6}{5}(x-5y-11)

Which is the required euation of tangent plane.

4 0
3 years ago
4+8/2* (6-3). Please answerrrr
Andrew [12]

Answer:

16

Step-by-step explanation:

brainliest plss

5 0
3 years ago
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Help me please!!!!!!!!
kompoz [17]

Answer:

The answer is A

Mean: 2

Median: 1

Step-by-step explanation:

3 0
2 years ago
(15pts) factor each completly
Natali [406]
Your answers:

10) (m+7)(m+8) 11) (b-10)(b+2)

12) (k-4)(k+10) 13) (n-9)(n-8)

14) (7r+3)(r-2) 15) (5a+6)(a+2)

Hope it helps
5 0
3 years ago
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Write the equation of a line that passes through the points (-1, -4) and (1, 3).
Elan Coil [88]

Answer:

y=3.5x-.5

Step-by-step explanation:

5 0
3 years ago
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