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MatroZZZ [7]
2 years ago
12

Can somebody help me?

Mathematics
1 answer:
GarryVolchara [31]2 years ago
3 0

Answer:

Step-by-step explanation: vcdvv

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Which equation is the inverse of 5y+4 = (x+3)^2 + 1/2
Anna35 [415]
C. -5y -4 = -(x+3)^2 -1/2

Hope I am right.
3 0
3 years ago
Read 2 more answers
Use lagrange multiplier techniques to find the local extreme values of f(x, y) = x2 − y2 − 2 subject to the constraint x2 + y2 =
dexar [7]
Given f(x,\ y)=x^2-y^2-2 subject to the constraint x^2+y^2=16

Let g(x,\ y)=x^2+y^2.

The gradient vectors of f and g are:

\nabla f(x,\ y)=\langle2x,-2y\rangle and \nabla g(x,\ y)=\langle2x,2y\rangle

By Lagrange's theorem, there is a number \lambda, such that

\langle2x,-2y\rangle=\lambda\langle2x,2y\rangle=\langle2\lambda x,2\lambda y\rangle

\lambda=\pm1

It can be seen that f(x,\ y)=x^2-y^2-2 has local extreme values at the given region.
6 0
3 years ago
How do I solve this? It says solve each right angle and then to round answers to the nearest tenth.
mariarad [96]

Answer:

19.2

Step-by-step explanation:

Add 12 squared to 15 squared(144+225)

Square root the answer(\sqrt{369})

Round to the nearest tenth(19.209)

7 0
3 years ago
8.
Alenkinab [10]

Answer:

17/5

Step-by-step explanation:

23=5x+6

23-6=5x

17=5x

X=17/5

6 0
3 years ago
PLEASE HELP!!! I am struggling
cestrela7 [59]

The ball will be at height 19 feet first at 0.7 sec and then at 1.42 sec.

<u>Explanation:</u>

We need to find the time at which the ball will be at height 19 feet.

Equation:

h = 3 + 34t - 16t²

19 = 3 + 34t - 16t²

16 = -16t² + 34t

-16t² + 34t - 16 = 0

On solving the equation, we get

t1 = 0.7 s and t2 = 1.42s

Therefore, the ball will be at height 19 feet first at 0.7 sec and then at 1.42 sec.

4 0
3 years ago
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