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emmasim [6.3K]
3 years ago
12

How many solutions does the equation have?

Mathematics
2 answers:
timurjin [86]3 years ago
7 0
4(2x+3)=2(3x-4)\\
8x+12=6x-8\\
2x=-20\\
x=-10 \Rightarrow A
creativ13 [48]3 years ago
6 0
A. One Solution.
If you distribute the 4 and 2 you're left with
8x+12=6x-8
+8 to both sides
8x+20=6x
-8x both sides
20=-2x
Divide by -2
X=-10
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Let f(x,y)=xex2−y and P=(9,81). (a) Calculate ∥∇fP∥. (b) Find the rate of change of f in the direction ∇fP. (c) Find the rate of
natali 33 [55]

Answer:

a) \sqrt[]{163^3+9^2}

b) \sqrt[]{163^3+9^2}

c) \sqrt[]{163^3+9^2}\cdot \frac{\sqrt[]{2}}{2}

Step-by-step explanation:

The given function is f(x) = xe^{x^2-y}. Recall the following:

-\nabla f = (\frac{df}{dx}, \frac{df}{dy} (The gradient of f is defined as the vector whose components are the partial derivatives of the function with respect to each of its variables)

- Given a direction vector v, that is a vector that is unitary, the rate of change of the function f in the direction v is given by

\nabla f \cdot v

- Recall that given two vectors a and b, the dot product between them is given by

a\cdot b = ||a|| ||b|| \cos(\theta)

- REcall that given a vector x, then x \cdot x = ||x||^2

where theta is the angle between both vectors and ||a|| is the norm of the vector a

- Given a vector  of components (x,y) its norm is given by \sqrt[]{x^2+y^2}.

a)Let us calculate first the gradient of f and the calculate it at the given point. We will omit the inner steps of derivation, so you must check that the gradient of f is given by

\nabla f = ((2x^2+1)e^{x^2-y},-xe^{x^2-y}). Since at P we have x=9, and y=81 the desired gradient is

\nabla f = (163,-9) and so the norm of the gradient at P is \sqrt[]{163^2+9^2}.

b) We want an unitary vector v for the gradient of f, so we take the gradient and divide it by its norm (i.e \frac{\nabla f}{||\nabla f||})

Hence, the rate of change is given by

\nabla f \cdot \frac{\nabla f}{||\nabla f||} = \frac{||\nabla f||^2}{||\nabla f||}=||\nabla f||

c). We are given that \theta = \pm45 ^\circ. We consider a vector a that is unitary, hence, the rate of change of f in the direction of vector a is given by

\nabla f \cdot a = ||a||||\nabla f||\cos(\pm 45 ^\circ) = \frac{\sqrt[]{2}}{2}||\nabla f||

3 0
4 years ago
Convert 220 from degrees to radians
liubo4ka [24]
220 degrees = 3.83972

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</span><span>I hope this helps.</span>
5 0
4 years ago
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4 0
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MrRa [10]
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Answer:

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