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2 years ago

9

A quadrilateral has no pairs of parallel sides. What two shapes could it be? (Not trapezoid/trapezium,concave quadrilateral conv

ex quadrilateral,kite,rhombus)

1 answer:

Nezavi [6.7K]2 years ago

7 0

Answer: kite and concave quadrilateral

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If ƒ={(5, 1),(6, 2),(7, 3),(8, 1),(9, 7)}, then the range of ƒ is {(1, 5),(2, 6),(3, 7),(1, 8),(7, 9)} {1, 2, 3, 7} {5, 6, 7, 8,

julsineya [31]

You need to just rewrite this

6 0

3 years ago

The sum of the digits of a two-digit number is 10. When the digits are reversed, the new number is 18 less than the original num

eimsori [14]

<h3>Answer:</h3>

64

<h3>Explanation:</h3>

6+4= 10

Then reverse the digits of 64, and you have 46

64-46= 18

3 0

3 years ago

Can someone help with numbers 1 and 2 for both of them tell me the answer and how you get it please I need the two of them do no

algol [13]

Answer:

look it up

Step-by-step explanation:

L.O.O.K I.T U.P.

5 0

3 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

James has $32 and earns $10 per week for his allowance. What is the initial value for the scenario described?

nignag [31]

32 is your initial value, because that is what he starts with

6 0

3 years ago

Read 2 more answers