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zhenek [66]
3 years ago
8

The mapping diagram shows a functional relationship. Complete the statement. f(3) is ___

Mathematics
2 answers:
klio [65]3 years ago
7 0

Answer:

-1 buddy.

u have just see what number it is going on.

AVprozaik [17]3 years ago
3 0

Answer:

f(3) is - 1

Step-by-step explanation:

f(3) means the function value when x = 3

3 is in the domain, the function value is in the range.

Thus when x = - 3 → y = - 1

f(3) is - 1

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What is 7.884 rounded to the nearest hundredth
Korvikt [17]

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The Answer is <u><em>7.88</em></u>

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8 0
3 years ago
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How do you do <br> 24 divided by 0.012
MariettaO [177]

Answer:

24 ÷ 0.012 =2000

Step-by-step explanation:

24 ÷ 0.012

= 2000

6 0
2 years ago
A computer training institute has 625 students that are paying a course fee of $400. Their research shows that for every $20 red
Anastasy [175]
1.
If no changes are made, the school has a revenue of :

625*400$/student=250,000$

2.
Assume that the school decides to reduce n*20$.

This means that there will be an increase of 50n students.

Thus there are 625 + 50n students, each paying 400-20n dollars.

The revenue is: 

(625 + 50n)*(400-20n)=12.5(50+n)*20(20-n)=250(n+50)(20-n)

3.

check the options that we have, 

a fee of $380 means that n=1, thus 

250(n+50)(20-n)=250(1+50)(20-1)=242,250   ($)


a fee of $320 means that n=4, thus

 250(n+50)(20-n)=250(4+50)(20-4)=216,000    ($)


the other options cannot be considered since neither 400-275, nor 400-325 are multiples of 20.

Conclusion, neither of the possible choices should be applied, since they will reduce the revenue.
6 0
3 years ago
Can anyone help me integrate :
worty [1.4K]
Rewrite the second factor in the numerator as

2x^2+6x+1=2(x+2)^2-2(x+2)-3

Then in the entire integrand, set x+2=\sqrt3\sec t, so that \mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt. The integral is then equivalent to

\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt

Note that by letting x+2=\sqrt3\sec t, we are enforcing an invertible substitution which would make it so that t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3} requires 0\le t or \dfrac\pi2. However, \tan t is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which \tan t>0 so that |\tan t|=\tan t. This allows us to write

=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt
=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt

We can show pretty easily that

\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C
\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C
\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C
\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C

which means the integral above becomes

=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C
=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C

Back-substituting to get this in terms of x is a bit of a nightmare, but you'll find that, since t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}, we get

\sec t=\dfrac{x+2}{\sqrt3}
\sec^2t=\dfrac{(x+2)^2}3
\tan t=\sqrt{\dfrac{x^2+4x+1}3}
\cot t=\sqrt{\dfrac3{x^2+4x+1}}
\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}
\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}

etc.
3 0
3 years ago
What is the value of x in the equation x – y = 30, when y = 15?
professor190 [17]


the answer is 45

because I know 15+15=30 and 30+15=45

so 45-15=30

hope it helps :)

3 0
3 years ago
Read 2 more answers
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