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Damm [24]
3 years ago
12

PLZ HELP ME What shape is this cross-section?

Mathematics
1 answer:
Irina18 [472]3 years ago
6 0

Answer:

i'm not sure if i'm right but think is cube

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Please guys help me. I will give you the brainliest if you give me the correct answers. ​
andrey2020 [161]

Answer:

vbvcfvbnbvcxcvbnvcxvbnmbvcvbvfdfgbhnbgvfgvbnbgfghbnbgvfgvbnbgfgbngfgbgfgfgfgfgfgfgfgfgfgfgrrrdrd

Step-by-step explanation:

3 0
3 years ago
Find equations of the tangent plane and the normal line to the given surface at the specified point. x + y + z = 8exyz, (0, 0, 8
Dima020 [189]

Let f(x,y,z)=x+y+z-8e^{xyz}. The tangent plane to the surface at (0, 0, 8) is

\nabla f(0,0,8)\cdot(x,y,z-8)=0

The gradient is

\nabla f(x,y,z)=\left(1-8yze^{xyz},1-8xze^{xyz},1-8xye^{xyz}\right)

so the tangent plane's equation is

(1,1,1)\cdot(x,y,z-8)=0\implies x+y+(z-8)=0\implies x+y+z=8

The normal vector to the plane at (0, 0, 8) is the same as the gradient of the surface at this point, (1, 1, 1). We can get all points along the line containing this vector by scaling the vector by t, then ensure it passes through (0, 0, 8) by translating the line so that it does. Then the line has parametric equation

(1,1,1)t+(0,0,8)=(t,t,t+8)

or x(t)=t, y(t)=t, and z(t)=t+8.

(See the attached plot; the given surface is orange, (0, 0, 8) is the black point, the tangent plane is blue, and the red line is the normal at this point)

4 0
3 years ago
Help I really don’t understand this
MAXImum [283]
Let x=3.77777777...
10x=37.777777777...
9x=10x-x=34
9x=34 so x=34/9=3 7/9
3 0
3 years ago
2 more questions thanks
sergey [27]
These are two questions and two answers.

1) Problem 17.

(i) Determine whether T is continuous at 6061.

For that  you have to compute the value of T at 6061 and the lateral limits of T when x approaches 6061.

a) T(x) = 0.10x if 0 < x ≤ 6061

T (6061) = 0.10(6061) = 606.1

b) limit of Tx when x → 6061.

By the left the limit is the same value of T(x) calculated above.

By the right the limit is calculated using the definition of the function for the next stage: T(x) = 606.10 + 0.18 (x - 6061)

⇒ Limit of T(x) when x → 6061 from the right = 606.10 + 0.18 (6061 - 6061) = 606.10

Since both limits and the value of the function are the same, T is continuous at 6061.

(ii) Determine whether T is continuous at 32,473.

Same procedure.

a) Value at 32,473

T(32,473) = 606.10 + 0.18 (32,473 - 6061) = 5,360.26

b) Limit of T(x) when x → 32,473 from the right

Limit = 5360.26 + 0.26(x - 32,473) = 5360.26

Again, since the two limits and the value of the function have the same value the function is continuos at the x = 32,473.

(iii) If  T had discontinuities, a tax payer that earns an amount very close to the discontinuity can easily approach its incomes to take andvantage of the part that results in lower tax.

2) Problem 18.

a) Statement Sk

You just need to replace n for k:

Sk = 1 + 4 + 7 + ... (3k - 2) = k(3k - 1) / 2

b) Statement S (k+1)

Replace

S(k+1) = 1 + 4 + 7 + ... (3k - 2) + [ 3 (k + 1) - 2 ] = (k+1) [ 3(k+1) - 1] / 2

Simplification:

1 + 4 + 7 + ... + 3k - 2+ 3k + 3 - 2] = (k + 1) (3k + 3 - 1)/2

                 k(3k - 1)/ 2 + (3k + 1) = (k + 1)(3k+2) / 2

Do the operations on the left side and  you will find it can be simplified to k ( 3k +1) (3 k + 2) / 2.

With that you find that the left side equals the right side which is a proof of the validity of the statement by induction.

4 0
2 years ago
Read 2 more answers
3x + 5 + 2x = 12 + 4x
Tatiana [17]
3x +5 +2x = 12 + 4x
3x+2x-4x =12 -5
x = 7
7 0
3 years ago
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