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Elenna [48]
3 years ago
9

What is the value of K? k = 8 o 4 k N 1 M

Mathematics
1 answer:
Vsevolod [243]3 years ago
7 0

Answer:

m

Step-by-step explanation:

it should be

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Factorizing as the Difference of Two Squares<br>1. 144x2 - 49​
igor_vitrenko [27]

Answer:

Factorizing as the Difference of Two Squares

1. 144x2 - 49​

Step-by-step explanation:

Where is the picture?

7 0
3 years ago
Please find the area of the square I forgot how to do this please help :(
Strike441 [17]

Answer:

85/2 or 42.5 ft²

Step-by-step explanation:

A = l x w

A = 7.5 (5 2/3)

= 15/2 (17/3)

= 255/6

= 85/2 ft² or 42.5 ft²

4 0
3 years ago
Read 2 more answers
Can anyone solve these 3 equations?
Nesterboy [21]

Answer:

1. x = 1, -2

2.  no solution

3.  x = 6 -4

hope this helps!

Step-by-step explanation:


3 0
3 years ago
Consider the Polynomial P(x)=x^3-5x^2-x+5. is (x-5) a factor?
Alborosie

Answer:

Option B, Yes, the remainder is 0, so x-5 is a factor of P(x)

Step-by-step explanation:

<u>Step 1:  Factor</u>

p(x) = x^3 - 5x^2 - x + 5

<em>p(x) = (x - 5)(x + 1)(x - 1)</em>

<em />

<em>Yes, the remainder is 0 so, x-5 is a factor of p(x)</em>

<em />

Answer:  Option B, Yes, the remainder is 0, so x-5 is a factor of P(x)

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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