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lana [24]
3 years ago
9

Hello I need help with concavity of graphs. I have included a picture of the 4 problems I would love to be checked to see if the

y are correct. My main questions are: what do you do when there's more than 2 concave up/down? And how do you write it if one of the points is open such as in letter A

Mathematics
1 answer:
Otrada [13]3 years ago
3 0

You would use the union symbol which is a U to glue the intervals together.

For instance, with problem 1 having two concave down intervals. You would write (-8,-4) U (0,4). Note that I'm using parenthesis only and no square brackets at all. The endpoints aren't included as they are open holes. We also don't include any point of inflection (POI). Your other answers are correct. Nice work.

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Find f(x) if it is known that f(x+3)=4x+12.
MArishka [77]

Solution:

we are given that

f(x+3)=4x+12

we have been asked to find f(x).

we can find the value of f(x) by re-writing the given function in the form of f(x+3) and thereafter we will replace (x+3) by x.

we can re-write the given function as below

f(x+3)=4(x+3)\\\\\Rightarrow f(x)=4x\\

Hence the expression representing f(x) is f(x)=4x+3.

8 0
2 years ago
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The space shuttle circled earth twenty times
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Answer:

498020 miles al

Step-by-step explanation:

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2 years ago
Factorise 3c^2-c<br><br><br> Please help me quick
Dafna11 [192]

Answer:

c(3c - 1)

Step-by-step explanation:

3c² - c

Common factor = c

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c(3c - 1)

5 0
3 years ago
4x^2 y+8xy'+y=x, y(1)= 9, y'(1)=25
jarptica [38.1K]

Answer with explanation:

\rightarrow 4x^2y+8x y'+y=x\\\\\rightarrow 8xy'+y(1+4x^2)=x\\\\\rightarrow y'+y\times\frac{1+4x^2}{8x}=\frac{1}{8}

--------------------------------------------------------Dividing both sides by 8 x

This Integration is of the form ⇒y'+p y=q,which is Linear differential equation.

Integrating Factor

 =e^{\int \frac{1+4x^2}{8x} dx}\\\\e^{\log x^{\frac{1}{8}+\frac{x^2}{2}}\\\\=x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}

Multiplying both sides by Integrating Factor  

x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\times [y'+y\times\frac{1+4x^2}{8x}]=\frac{1}{8}\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\\\\ \text{Integrating both sides}\\\\y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\frac{1}{8}\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx \\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=-[x^{\frac{9}{8}}]\times\frac{ \Gamma(0.5625, -x^2)}{(-x^2)^{\frac{9}{16}}}\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=(-1)^{\frac{-1}{8}}[ \Gamma(0.5625, -x^2)]+C-----(1)

When , x=1, gives , y=9.

Evaluate the value of C and substitute in the equation 1.

6 0
3 years ago
I really don't understand this honestly. Please help lol
klasskru [66]

Answer:

300

Step-by-step explanation:

I t is good

6 0
2 years ago
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