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mixer [17]
3 years ago
12

I WILL GIVE BRAINLIEST PLEASE ANSWER ASAP

Mathematics
1 answer:
Elena-2011 [213]3 years ago
7 0
Rectangular..................
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Which of the following are solutions to the equation below? check all that apply 4x^2-81=0
seropon [69]

Answer:

9/2

Step-by-step explanation:

4 {x}^{2}  = 81 \\  {x}^{2}  =  \frac{81}{4}  \\ x =  \frac{ \sqrt{81} }{ \sqrt{4} }  \\ x =  \frac{9}{2}

3 0
3 years ago
Read 2 more answers
Complete the remainder of the
N76 [4]

Answer: 2, 4, 6, 8

Step-by-step explanation:

Just plug x into the equation for each one.

x = -3

y=\frac{2(-3)}{3} +4\\y=\frac{-6}{3} +4\\y=2

With this one, you can see it is a linear equation and for every increase of 3 on x, y in increased by 2.

x   -6   -3   0   3   6

y   0   2    4   6   8

8 0
2 years ago
What type of function is represented in the table? x|0|1|2|3|4|5|
mezya [45]

Answer:

This may be wrong but im guessing it is B.logarithmic

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Find the equation of the line passing<br> through the points (2, 1) and (5, 10).<br> y = [? ]x + [ ]
sladkih [1.3K]

Answer:

y=3x-5

Step-by-step explanation:

m=10-1/5-2

m=9/3

m=3

y=3x+b

x=2, y=1

1=3(2)+b

1=6+b

-5=b

y=3x-5

8 0
3 years ago
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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