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4 years ago

Consider functions f and g

Mathematics

1 answer:

Ymorist [56]4 years ago

7 0

Answer: B

f(g(x)) = (3x - 1)² + 7(3x - 1) = 9x² + 15x - 6

Step-by-step explanation:

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Y=2x/a+2b solve for a

mina [271]

Solve for a.

y=2x/a+2b

Multiply both sides by a.

ay=2ab+2x

Add -2ab to both sides.

ay+−2ab=2ab+2x+−2ab

−2ab+ay=2x

Factor out variable a.

a(−2b+y)=2x

Divide both sides by -2b+y.

a(−2b+y) / −2b+y = 2x / −2b+y

a = 2x / −2b+y

6 0

3 years ago

Indicate a general rule for the nh term of this sequence.<br> 12a, 15a, 18a, 21a, 24a,...

adelina 88 [10]

Answer:

The general rule for the nth term of this sequence will be:

$a_n=3na+9a$

Step-by-step explanation:

Given the sequence

12a, 15a, 18a, 21a, 24a,...

An arithmetic sequence has a constant difference 'd' and is defined by

$a_n=a_1+\left(n-1\right)d$

Here,

a₁ = 12a

computing the differences of all the adjacent terms

d = 15a-12a = 3a, d = 18a-15a=3a, d=21a-18a=3a, d=24a-21a=3a

using the nth term formula

$a_n=a_1+\left(n-1\right)d$

substituting a₁ = 12a, d = 3a

$a_n=12a+\left(n-1\right)3a$

$=12a+3na-3a$

$=3na+9a$

Therefore, the general rule for the nth term of this sequence will be:

$a_n=3na+9a$

6 0

3 years ago

Solve the equation using the substitution u = y/x. When u = y/x is substituted into the equation, the equation becomes separable

bekas [8.4K]

Answer:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

Step-by-step explanation:

First step: I'm going to solve our substitution for y:

$u=\frac{y}{x}$

Multiply both sides by x:

$ux=y$

Second step: Differentiate the substitution:

$u'x+u=y'$

Third step: Plug in first and second step into the given equation dy/dx=f(x,y):

$u'x+u=\frac{x(ux)+(ux)^2}{3x^2+(ux)^2}$

$u'x+u=\frac{ux^2+u^2x^2}{3x^2+u^2x^2}$

We are going to simplify what we can.

Every term in the fraction on the right hand side of equation contains a factor of $x^2$ so I'm going to divide top and bottom by $x^2$ :

$u'x+u=\frac{u+u^2}{3+u^2}$

Now I have no idea what your left hand side is suppose to look like but I'm going to keep going here:

Subtract u on both sides:

$u'x=\frac{u+u^2}{3+u^2}-u$

Find a common denominator: Multiply second term on right hand side by $\frac{3+u^2}{3+u^2}$ :

$u'x=\frac{u+u^2}{3+u^2}-\frac{u(3+u^2)}{3+u^2}$

Combine fractions while also distributing u to terms in ( ):

$u'x=\frac{u+u^2-3u-u^3}{3+u^2}$

$u'x=\frac{-u^3+u^2-2u}{3+u^2}$

Third step: I'm going to separate the variables:

Multiply both sides by the reciprocal of the right hand side fraction.

$u' \frac{3+u^2}{-u^3+u^2-2u}x=1$

Divide both sides by x:

$\frac{3+u^2}{-u^3+u^2-2u}u'=\frac{1}{x}$

Reorder the top a little of left hand side using the commutative property for addition:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

The expression on left hand side almost matches your expression but not quite so something seems a little off.

5 0

3 years ago

This is a classic joke question. Only the real ones know the answer.<br> what is 10+9

Karo-lina-s [1.5K]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Can i get some help with #5 and below thankyou

trasher [3.6K]

Answer:

Step-by-step explanation:

what the other guy said!

6 0

3 years ago