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Alekssandra [29.7K]
2 years ago
8

Please answer asap (look at picture) please answer both

Mathematics
1 answer:
exis [7]2 years ago
3 0
0,2 is the answer since that’s where both lines intersect
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See file attached below:
pochemuha

Answer:

Goodluck my dude.

Step-by-step explanation:

7 0
2 years ago
Y-intercept:<br>Rate:<br>Growth or Decay?​
Viefleur [7K]

Answer:

Growth

Step-by-step explanation:

3 0
3 years ago
lim x rightarrow 0 1 - cos ( x2 ) / 1 - cosx The limit has to be evaluated without using l'Hospital'sRule.
zaharov [31]

Answer with Step-by-step explanation:

Given

f(x)=\frac{1-cos(2x)}{1-cos(x)}\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-(cos^2{x}-sin^2{x})}{1-cos(x)})\\\\(\because cos(2x)=cos^2x-sin^2x)\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-cos^2x}{1-cos(x)}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}(\frac{(1-cosx)(1+cosx)}{1-cosx}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}((1+cosx)+\frac{sin^2x}{1-cosx})\\\\\therefore \lim_{x \rightarrow 0}f(x)=1

6 0
3 years ago
Determine whether or not the vector field is conservative. if it is conservative, find a function f such that f = ∇f. (if the ve
Akimi4 [234]
A three-dimensional vector field is conservative if it is also irrotational, i.e. its curl is \mathbf 0. We have

\nabla\times\mathbf f(x,y,z)=-2e^{-x}\,\mathbf k

so this vector field is not conservative.

- - -

Another way of determining the same result: We want to find a scalar function f(x,y,z) such that its gradient is equal to the given vector field, \mathbf f(x,y,z):

\nabla f(x,y,z)=\mathbf f(x,y,z)

For this to happen, we need to satisfy

\begin{cases}f_x=ye^{-x}\\f_y=e^{-x}\\f_z=2z\end{cases}

From the first equation, integrating with respect to x yields

f_x=ye^{-x}\implies f(x,y,z)=-ye^{-x}+g(y,z)

Note that g *must* be a function of y,z only.

Now differentiate with respect to y and we have

f_y=-e^{-x}+g_y=e^{-x}\implies g_y=2e^{-x}\implies g(y,z)=2ye^{-x}+\cdots

but this contradicts the assumption that g(y,z) is independent of x. So, the scalar potential function does not exist, and therefore the vector field is not conservative.
8 0
3 years ago
A. Write the linear system as a matrix equation
umka21 [38]
AX = B

1 -1 -1 x 0
0 -2 -1 y = -2
2 1 0 z -4

x
y = A^-1 B
z

x -1 1 1 0 6
y = 2 -2 -1 . 2 = -8
z -4 3 2 4 14





7 0
2 years ago
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