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Lady_Fox [76]
3 years ago
6

Can anyone help me please

Mathematics
1 answer:
denis23 [38]3 years ago
6 0
Im pretty sure that it is D or the last one. If you're equally sharing something, you would have to divide it.
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The mean of the new data set is 2.76, and the range of the new data set is 1.5.
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PLS HELP FIRST ONE RIGHT GETS BRAINLEST |||| A produce stand is packing blueberries into 2/5 pound containers. How many containe
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Step-by-step explanation:

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Consider the function <br>f(x) = 8x^3 +1.<br><br> If f(g(x)) = x and g(f(x)) = x , find g(x) ?
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g( \: f(x) \: ) = x

And

f( \: g(x) \: ) = x_________________________________

So :

g(x) =  {f}^{ - 1}(x)

And

f(x) =  {g}^{ - 1}(x)

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So to find g(x) , we must find the inverse of f(x) .

Let's do it .....

f(x) = 8 {x}^{3} + 1

y = 8 {x}^{3} + 1

Subtract the sides of the equation minus 1

y - 1 = 8 {x}^{3}

Divided the sides of the equation by 8

\frac{y - 1}{8} =  {x}^{3} \\

From the sides of the equation, we take the radical with interval 3

\sqrt[3]{ \frac{y - 1}{8} } = x \\

\sqrt[3]{ \frac{1}{8}(y - 1) } = x \\

\frac{1}{2} \sqrt[3]{y - 1} = x \\

\frac{ \sqrt[3]{y - 1} }{2} = x \\

So ;

{f}^{ - 1}(x) = \frac{ \sqrt[3]{x - 1} }{2} \\

Now we find g(x) which is :

g(x) =  \frac{ \sqrt[3]{x - 1} }{2} \\

_________________________________

And we're done.

Thanks for watching buddy good luck.

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7 0
3 years ago
Without solving an equation how can you determine if the equation has one solution?
amid [387]

Answer:

Yes. If it has one variable and not exponential the solution is 1.

If is one variable and to power of 2 it is quadratic and two solutions.

If it two variable with no power it is simultaneous and with 2 solutions. E.t.c

Step-by-step explanation:

7 0
4 years ago
Suppose ABCD is a convex quadrilateral. Points M and N are midpoints of sides BC and AD, respectively. The area of the quadrilat
Katena32 [7]

Answer:

12 unit²

Step-by-step explanation:

area ΔAMC = 1/2 CM x AH' = 1/4 BC x AH'

area ΔABM = 1/2 BM x AH' = 1/4 BC x AH'

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for the same calculation, we can prove

area ΔACN = area ΔDCN

area ΔAMC + area ΔACN = 6

area ΔABM + area ΔDCN = 6

ABCD = area ΔAMC + area ΔACN + area ΔABM + area ΔDCN = 6 = 6 = 12

7 0
3 years ago
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