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notsponge [240]
4 years ago
12

Solve x^2-x-42=0 for quadratic equation​

Mathematics
2 answers:
harkovskaia [24]4 years ago
6 0

Answer:x

x=7,-6

Step-by-step explanation:

Solve the equation for  x  by finding  a ,  b , and  c  of the quadratic then applying the quadratic formula.

ps. can you mark me brainliest if i got it right??? plz? thanks

11Alexandr11 [23.1K]4 years ago
6 0

Answer:  x = 7 , -6

Step-by-step explanation: Solve the equation for x by finding a, b and c of the quadric then applying the quadric formula.

Hope this helps you out! ☺

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12 inches to 36 inches
Schach [20]
You have to Multiply times 3...
12•3=36
36/12=3
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3 years ago
Look at this set of ordered pairs:
jeka57 [31]

Answer:

  • No

Step-by-step explanation:

<u>Given ordered pairs:</u>

  • (3, -13)
  • (3, 2)
  • (18,-5)

There is a repeat of domains (3) with different range (-13 and 2) but the function can't have repeat domains so it is not a function

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I have a question for someone out there I'm 10 and need some help what is 1/10+10/100​
Ratling [72]

Answer:

0.2

Step-by-step explanation:

1/10 = 0.1

10/100 = 0.1

0.1+0.1 = 0.2

7 0
3 years ago
Some more free-points education is important
makkiz [27]

Answer:

Thank you for the free-points!!! <3

Step-by-step explanation:

5 0
3 years ago
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Show that the line integral is independent of path by finding a function f such that ?f = f. c 2xe?ydx (2y ? x2e?ydy, c is any p
Juli2301 [7.4K]
I'm reading this as

\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy

with \nabla f=(2xe^{-y},2y-x^2e^{-y}).

The value of the integral will be independent of the path if we can find a function f(x,y) that satisfies the gradient equation above.

You have

\begin{cases}\dfrac{\partial f}{\partial x}=2xe^{-y}\\\\\dfrac{\partial f}{\partial y}=2y-x^2e^{-y}\end{cases}

Integrate \dfrac{\partial f}{\partial x} with respect to x. You get

\displaystyle\int\dfrac{\partial f}{\partial x}\,\mathrm dx=\int2xe^{-y}\,\mathrm dx
f=x^2e^{-y}+g(y)

Differentiate with respect to y. You get

\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]
2y-x^2e^{-y}=-x^2e^{-y}+g'(y)
2y=g'(y)

Integrate both sides with respect to y to arrive at

\displaystyle\int2y\,\mathrm dy=\int g'(y)\,\mathrm dy
y^2=g(y)+C
g(y)=y^2+C

So you have

f(x,y)=x^2e^{-y}+y^2+C

The gradient is continuous for all x,y, so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is

\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy=f(4,1)-f(1,0)=\frac9e
8 0
3 years ago
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