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Sergeeva-Olga [200]
2 years ago
12

How to do linear functions using the table ?

Mathematics
1 answer:
Ira Lisetskai [31]2 years ago
3 0

Answer:

-8 = m-2+b

-2 = m-0+b

4 = m2+b

10 = m4+b

Step-by-step explanation:

You just put them in y = mx+b format

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4x^(4)+8x^(3)-2x^(2) write as a product
True [87]
Answer:
764x
Step by step explanation:
4^4=256
8^3=512
2^2=4
256-4+512=764

765x
4 0
1 year ago
If the interest of a sum at 7.5% p.a in 4 years is rs 1050, find the sum​
Alborosie

formula p= 100 I/RT

100×1050÷7.5×4

105000÷30

p= sh 3500

6 0
3 years ago
I need answer Immediately pls!!!!!!!
Anuta_ua [19.1K]

Answer:

1/14

Step-by-step explanation:

There is only 1 common multiple of 4 and 6 between 1 and 14.

So the probability is:

P = \frac{1}{14}

8 0
3 years ago
Read 2 more answers
A horse's mane grows at the rate of 6/5 centimeters per month. How many months will it take the horse's mane to grow 24/10 centi
Fittoniya [83]
24/10
---------
6/5

2 months
6 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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