3 years ago

Is F=((x,y) | x+y=10 a function or a relation

Mathematics

1 answer:

asambeis [7]3 years ago

7 0

that is a function man

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Find the value p with working

drek231 [11]

Answer:

64/672/882(986)*866772

5 0

2 years ago

Read 2 more answers

Pls help! I need help with a and c I did b already!

laila [671]

Answer:

A= y= -2x+7.

B= y= - $\frac{3}{2}$ x+ 6

Step-by-step explanation:

The linear equation is y=mx+b

A= y= -2x+7.

-2 represents he slope and 7 represents the y intercept.

B= y= - $\frac{3}{2}$ x+ 6

- $\frac{3}{2}$ is the slop and 6 is the y intercept.

Hope this helps :)

8 0

2 years ago

Chang made $144 for 8 hours of work at the same rate how many hours would he have to work to make 234

olganol [36]

Answer:

ben daha küçük olduğum için özür dilerim yapamam 4. sınıfa gidiyorum

Step-by-step explanation:

VE puan kazanmam gerek

3 0

2 years ago

Can anyone help me with this please

tigry1 [53]

It’s b here is proof of what i got

4 0

2 years ago

Verify that P = Ce^t /1 + Ce^t is a one-parameter family of solutions to the differential equation dP dt = P(1 − P).

NemiM [27]

Answer:

See verification below

Step-by-step explanation:

We can differentiate P(t) respect to t with usual rules (quotient, exponential, and sum) and rearrange the result. First, note that

$1-P=1-\frac{ce^t}{1+ce^t}=\frac{1+ce^t-ce^t}{1+ce^t}=\frac{1}{1+ce^t}$

Now, differentiate to obtain

$\frac{dP}{dt}=(\frac{ce^t}{1+ce^t})'=\frac{(ce^t)'(1+ce^t)-(ce^t)(1+ce^t)'}{(1+ce^t)^2}$

$=\frac{(ce^t)(1+ce^t)-(ce^t)(ce^t)}{(1+ce^t)^2}=\frac{ce^t+ce^{2t}-ce^{2t}}{(1+ce^t)^2}=\frac{ce^t}{(1+ce^t)^2}$

To obtain the required form, extract a factor in both the numerator and denominator:

$\frac{dP}{dt}=\frac{ce^t}{1+ce^t}\frac{1}{1+ce^t}=P(1-P)$

3 0

3 years ago