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motikmotik
3 years ago
11

Find the equation for the tangent plane and the normal line at the point

Mathematics
1 answer:
hodyreva [135]3 years ago
7 0
X =9 had the samw question on sat

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What is the answer to:<br> 3(x-2)+2(4x-1)
irga5000 [103]
11x-8 is the answer
8 0
3 years ago
Read 2 more answers
Use the distributive property to simplify the expressions. 1B. -3(9x-q) 2B. 2(12+5p) 3b. -7(-8y+7) 4b. 10(2+7c) 5b. -8(7+11k) 6b
Debora [2.8K]
1B: -3(9x-q)
       -27x + 3q


Step 1: Distribute the -3 to the 9x to get -27x.
Step 2: Distribute the -3 to the -q.


2B: 2(12+5p)
       24 + 10p

Step 1: Distribute the 2 to the 12 to get 24.
Step 2: Distribute the 2 to the 5p to get 10p.

3B: -7(-8y+7)
       56y - 49


Step 1: Distribute the -7 to the -8y to get 56y.
Step 2: Distribute the -7 to the 7 to get -49.

4B: 10(2+7c)
        20 + 70c


Step 1: Distribute the 10 to the 2 to get 20. 
Step 2: Distribute the 10 to the 7c to get 70c.

5B: -8(7+11k)
       -56 - 88k


Step 1: Distribute the -8 to the 7 to get -56.
Step 2: Distribute the -8 to the 11k to get -88k.

6B: -(3-7u)
       -1(3-7u)
       -3 + 7u


Step 1: Place a 1 after the negative symbol to symbolize -1.
Step 2: Distribute the -1 to the three to get -3.
Step 3: Distribute the -1 to the -7u to get 7u.


7B: -6(5p + s)
       -30p - 6s

Step 1: Distribute the -6 to the 5p to get -30p.
Step 2: Distribute the -6 to the s to get -6s.

:) :D




8 0
3 years ago
Does anyone know the answer to this!?
PolarNik [594]

Answer:

|A_y|=\left|\begin{array}{cc}12&7\\ \\17&-51\end{array}\right|

Step-by-step explanation:

You can use these Cramer's formulas to solve for x and y:

x=\dfrac{|A_x|}{|A|},\\ \\y=\dfrac{|A_y|}{|A|},

where

|A|=\left|\begin{array}{cc}12&-13\\ \\17&-22\end{array}\right|\\ \\ \\|A_x|=\left|\begin{array}{cc}7&-13\\ \\-51&-22\end{array}\right|\\ \\ \\|A_y|=\left|\begin{array}{cc}12&7\\ \\17&-51\end{array}\right|

So,

|A_y|=\left|\begin{array}{cc}12&7\\ \\17&-51\end{array}\right|

4 0
3 years ago
|(a + b)x + y = 4<br> |x + (a - b)y + 10
leonid [27]

I don't know the answer but try using PEMDAS. (Parentheses, Exponents, Multiplication, Division, Addition, and then Subtraction)

4 0
2 years ago
Prove that if one solution for a quadratic equation of the form x 2 + bx + c = 0 is rational (where b and c are rational), then
erica [24]
x^2+bx+c = (x-r)(x-s)\\&#10;x^2+bx+c = x^2-(r+s)x+rs\\\\&#10;b = -(r+s)\\ c = rs

Say r is rational. Suppose for a second, that s is not. Then, r+s is irrational. But this contradicts the fact that b is rational.

So, if one root is rational, then the other root is also rational
7 0
3 years ago
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