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Nonamiya [84]
3 years ago
5

3x +4 and =-23 x=3and +1

Mathematics
1 answer:
Hunter-Best [27]3 years ago
6 0

Answer:

Im not sure if you're asking us to solve the function operation but if you are heres what I got:

3x + 4 and = -23 =

x = 3 =

1 =

Sorry if this is wrong

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Please help, my brain just sucks rn. The area of a rectangle is 93.6 square inches. If the length of one of its sides is 5.2 in.
snow_tiger [21]

46.4 inch

Step-by-step explanation:

Area=l×b

93.6 =5.2×b

93.6/5.2 =b

So b= 18

Again,

p=2(l+b)

p=2(18+5.2)

p= 2×23.2

p=46.4

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Why is not reasonable to say that 4.23 is less than 4.135
AysviL [449]

Simply because 4.135 is a larger number than 4.23.

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Which two numbers on the number line have an absolute value of 2.75?
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Answer:

2.75 and -2.75

Step-by-step explanation:

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Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 6x(1/3) + 3x(4/3). You must justi
stealth61 [152]
Applying our power rule gets us our first derivative,

\rm f'(x)=6\frac13x^{-2/3}+3\cdot\frac43x^{1/3}

simplifying a little bit,

\rm f'(x)=2x^{-2/3}+4x^{1/3}

looking for critical points,

\rm 0=2x^{-2/3}+4x^{1/3}

We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.

0=2x^{-2/3}\left(1+2x\right)

When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.

Then apply your Zero-Factor Property,

\rm 0=2x^{-2/3}\qquad\qquad\qquad 0=(1+2x)

and solve for x in each case to find your critical points.

Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.

Let's come back to this,

\rm f'(x)=2x^{-2/3}+4x^{1/3}

and take our second derivative.

\rm f''(x)=-\frac43x^{-5/3}+\frac43x^{-2/3}

Looking for inflection points,

\rm 0=-\frac43x^{-5/3}+\frac43x^{-2/3}

Again, pulling out the smaller power of x, and fractional part,

\rm 0=-\frac43x^{-5/3}\left(1-x\right)

And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.

Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.
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