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velikii [3]
3 years ago
12

What do you do in this problem

Mathematics
2 answers:
frez [133]3 years ago
6 0

Answer:

6 seconds to 10 mph would be 2.6

Step-by-step explanation:

Citrus2011 [14]3 years ago
5 0
So it took him 6 seconds to get to 10 mph

10/6 = 1 2/3 so he is accelerating at a pace of 1 and 2/3 miles per second

Hope this helps
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Is (4,-5), (0,-5), (11,30), (17,10) a function
Ratling [72]

Answer:

Yes, it is a function

Step-by-step explanation:

A function is a set of ordered pairs where there is no two of the same domain. The domain is the first coordinate which is x. In this case, the x coordinates are 4, 0, 11, 17. There is no two of the same domain so this is a function.

Hope This Helps :)

4 0
3 years ago
You and your friends are going to a local concert. Each ticket costs $14. The total amount of ticket sales for the concert was $
serious [3.7K]

Answer: 628

Step-by-step explanation:

8792/14 = 628

5 0
4 years ago
Determine whether the stated conclusion is valid based on the given information. Explain your reasoning.
Oduvanchick [21]
The answer is D. The law of detachment states that
"If P than Q. P."
In this case passing the bar exam is P and getting to practice law is Q. So if we know that Candice is allowed to practice law if she passes the bar exam and that she did pass the exam. The conclusion that Candice can now practice law is valid.
4 0
3 years ago
WILL GIVE BRAINLST <br> ANSWER PLSS
Gelneren [198K]
The answer should be C
3 0
3 years ago
Read 2 more answers
Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0
Leokris [45]

Substitute v(x)=-2x+y(x), so that \dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}. Then the ODE is equivalent to

\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7

which is separable as

\dfrac{\mathrm dv}{v^2-9}=\mathrm dx

Split the left side into partial fractions,

\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)

so that integrating both sides is trivial and we get

\dfrac{\ln|v-3|-\ln|v+3|}6=x+C

\ln\left|\dfrac{v-3}{v+3}\right|=6x+C

\dfrac{v-3}{v+3}=Ce^{6x}

\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}

\dfrac6{v+3}=1-Ce^{6x}

v=\dfrac6{1-Ce^{6x}}-3

-2x+y=\dfrac6{1-Ce^{6x}}-3

y=2x+\dfrac6{1-Ce^{6x}}-3

Given the initial condition y(0)=0, we find

0=\dfrac6{1-C}-3\implies C=-1

so that the ODE has the particular solution,

\boxed{y=2x+\dfrac6{1+e^{6x}}-3}

5 0
3 years ago
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