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Pachacha [2.7K]
3 years ago
10

Solve for x - x + 2 = x - 5 - 3x

Mathematics
1 answer:
storchak [24]3 years ago
7 0
The answer is x=-7 you just have to cancel the like factors out n you don’t have to divide nothing in this equation.
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Which mathematical concepts did Euler's legacy include? Check all that apply.
quester [9]

Answer:

a)the notation for the imaginary unit

c) the formalization of a function notation

e) the notation for the base of the natural logarithm

Step-by-step explanation:

it said i was correct :P

7 0
3 years ago
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AB<br> C D<br> E<br> F +3<br> x2 -3x -18
madam [21]

Answer:

(x + 3) ( x - 6)

Step-by-step explanation:

x² - 3x - 18

x² - 6x + 3x - 18

x(x - 6) + 3(x - 6)

(x + 3) ( x - 6)

5 0
3 years ago
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Find the value of this expression if x=9 x2 + 6 / x - 6
GREYUIT [131]

Answer:

87/3

Step-by-step explanation:

Here we are given the expression:

\frac{x^{2}+6}{x-6}

Now we have to find the value of this expression when x equals 9.

So plugging the value of x as -9 in the given expression:

\frac{x^{2}+6}{x-6}

=\frac{(9)^{2}+6}{9-6}

=87/3

So on evaluating we get the value of expression as 87/3.



3 0
3 years ago
Read 2 more answers
Add<br> 6 3/5 + (-6 1/3<br> Enter your answer as a simplified fraction in the box
Sholpan [36]
Okay so basically the answer would be 4/15
8 0
2 years ago
Find the maximum and minimum values of the function f(x,y)=2x2+3y2−4x−5 on the domain x2+y2≤100. The maximum value of f(x,y) is:
ryzh [129]

First find the critical points of <em>f</em> :

f(x,y)=2x^2+3y^2-4x-5=2(x-1)^2+3y^2-7

\dfrac{\partial f}{\partial x}=2(x-1)=0\implies x=1

\dfrac{\partial f}{\partial y}=6y=0\implies y=0

so the point (1, 0) is the only critical point, at which we have

f(1,0)=-7

Next check for critical points along the boundary, which can be found by converting to polar coordinates:

f(x,y)=f(10\cos t,10\sin t)=g(t)=295-40\cos t-100\cos^2t

Find the critical points of <em>g</em> :

\dfrac{\mathrm dg}{\mathrm dt}=40\sin t+200\sin t\cos t=40\sin t(1+5\cos t)=0

\implies\sin t=0\text{ OR }1+5\cos t=0

\implies t=n\pi\text{ OR }t=\cos^{-1}\left(-\dfrac15\right)+2n\pi\text{ OR }t=-\cos^{-1}\left(-\dfrac15\right)+2n\pi

where <em>n</em> is any integer. We get 4 critical points in the interval [0, 2π) at

t=0\implies f(10,0)=155

t=\cos^{-1}\left(-\dfrac15\right)\implies f(-2,4\sqrt6)=299

t=\pi\implies f(-10,0)=235

t=2\pi-\cos^{-1}\left(-\dfrac15\right)\implies f(-2,-4\sqrt6)=299

So <em>f</em> has a minimum of -7 and a maximum of 299.

4 0
3 years ago
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