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3 years ago

What substitution Should be used to re-write 26(x^3+1)^2-22(x^3+1)-3=0 as a quadratic function?

Mathematics

1 answer:

mel-nik [20]3 years ago

7 0

Answer:

$x^3+1$

Step-by-step explanation:

$26(x^3+1)^2-22(x^3+1)-3=0$

Comparing to

$A(u)^2+B(u)+C =0$

Where A,B,C are constants

You should see that we need to substitute the $x^3+1$ with u.

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Find the nth term of

kykrilka [37]

Answer:

Step-by-step explanation:

17,20,23,26,29,31,34,37,40,43,46,49,51,54,57,60,63,66,69,71,74,77,80,83,86,

89,91,94,97,100

4 0

2 years ago

Put these numbers in order from least to greatest 1/8, -0.333, -1/7

Gekata [30.6K]

Answer:

-0.333, -1/7, 1/8

Step-by-step explanation:

1/8 is the greatest because it is a positive number while the other two numbers are negative.

-0.333 or -1/3 is smaller than -1/7.

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1 year ago

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lina2011 [118]

Answer:

2x + 4 I would assume

3 + 1 = 4

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2 years ago

Read 2 more answers

On their first day of vacation, a family spent $127.92 for meals. The family spent $160.01 for meals the second day. By what amo

ehidna [41]

Answer:

D=32.09

Step-by-step explanation:

160.01-127.92=32.09

5 0

3 years ago

Read 2 more answers

Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z

kompoz [17]

$f_{X,Y,Z}(x,y,z)=\begin{cases}Ce^{-(0.5x+0.2y+0.1z)}&\text{for }x\ge0,y\ge0,z\ge0\\0&\text{otherwise}\end{cases}$

is a proper joint density function if, over its support, $f$ is non-negative and the integral of $f$ is 1. The first condition is easily met as long as $C\ge0$ . To meet the second condition, we require

$\displaystyle\int_0^\infty\int_0^\infty\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=100C=1\implies \boxed{C=0.01}$

b. Find the marginal joint density of $X$ and $Y$ by integrating the joint density with respect to $z$ :

$f_{X,Y}(x,y)=\displaystyle\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dz=0.01e^{-(0.5x+0.2y)}\int_0^\infty e^{-0.1z}\,\mathrm dz$

$\implies f_{X,Y}(x,y)=\begin{cases}0.1e^{-(0.5x+0.2y)}&\text{for }x\ge0,y\ge0\\0&\text{otherwise}\end{cases}$

Then

$\displaystyle P(X\le1.375,Y\le1.5)=\int_0^{1.5}\int_0^{1.375}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy$

$\approx\boxed{0.12886}$

c. This probability can be found by simply integrating the joint density:

$\displaystyle P(X\le1.375,Y\le1.5,Z\le1)=\int_0^1\int_0^{1.5}\int_0^{1.375}f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz$

$\approx\boxed{0.012262}$

7 0

3 years ago