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3 years ago

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

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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

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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:

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3 years ago

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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