Ad

Hi,

g(1) = 2(1) - 3 = -1

So f(g(1)) = f(-1) = (-1)² + 2 = 1 + 2 = 3

f(g(1)) = 3 <==ANSWER

A second way:

f(g(x)) = (2x - 3)² + 2 =

4x² - 12x + 9 + 2 =

4x² - 12x + 11

f(g(1)) = 4(1)² - 12(1) + 11 = 4 - 12 + 11 = 3

f(g(1)) = 3

There you go...

Posted on Aug 08, 2010

Ad

Hi there,

Save hours of searching online or wasting money on unnecessary repairs by talking to a 6YA Expert who can help you resolve this issue over the phone in a minute or two.

Best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.

Here's a link to this great service

Good luck!

Posted on Jan 02, 2017

Ad

easy:

2log(x) + log(3/4) - log(2x + 3/4) = 0

log(x^2) + log(3/4) = log(2x + 3/4)

log(3/4*x^2) = log(2x + 3/4)

3/4*x^2 = 2x + 3/4

multiply equation by 4:

3x^2 = 8x + 3

3x^2 - 8x - 3 = 0

solve this quadratic eq. and you get

x1 = 3

x2 = -1/3

does it make sense?

2log(x) + log(3/4) - log(2x + 3/4) = 0

log(x^2) + log(3/4) = log(2x + 3/4)

log(3/4*x^2) = log(2x + 3/4)

3/4*x^2 = 2x + 3/4

multiply equation by 4:

3x^2 = 8x + 3

3x^2 - 8x - 3 = 0

solve this quadratic eq. and you get

x1 = 3

x2 = -1/3

does it make sense?

Mar 05, 2017 | Computers & Internet

If all sides are equal then 2x + 5 = 4x - 3

2x = 8

x = 4

so the perimeter is 3 times either expression above

3 * (8 + 5) = 39

.

2x = 8

x = 4

so the perimeter is 3 times either expression above

3 * (8 + 5) = 39

.

Jan 12, 2017 | The Computers & Internet

1) 2x + 5y = 7

2) 3x + 6y = 3

I'm going to use the method of elimination to solve for x and y.

Multiply 1) by 3 and 2) by 2 to allow the x's to be eliminated.

1) 6x + 15y = 21

2) 6x + 12y = 6

Now subtract line 2 from line 1.

0x + 3y = 15

---- ----

3 3 divide both sides by 3 to get y by itself.

y =5.

Substitute into 1) to calculate x.

2x + 5(5) = 7

2x + 25 = 7

2x + 25 -25 = 7 - 25

2x = -18

---- ----- divide both sides by 2 to get x by itself

2 2

x = -9

Check by plugging in answer into the other equation, in this case 2)

3 (-9) + 6(5) = 3

-27 + 30 = 3

3 = 3

We did it correctly and checked to prove that we did it right.

Good luck.

Paul

2) 3x + 6y = 3

I'm going to use the method of elimination to solve for x and y.

Multiply 1) by 3 and 2) by 2 to allow the x's to be eliminated.

1) 6x + 15y = 21

2) 6x + 12y = 6

Now subtract line 2 from line 1.

0x + 3y = 15

---- ----

3 3 divide both sides by 3 to get y by itself.

y =5.

Substitute into 1) to calculate x.

2x + 5(5) = 7

2x + 25 = 7

2x + 25 -25 = 7 - 25

2x = -18

---- ----- divide both sides by 2 to get x by itself

2 2

x = -9

Check by plugging in answer into the other equation, in this case 2)

3 (-9) + 6(5) = 3

-27 + 30 = 3

3 = 3

We did it correctly and checked to prove that we did it right.

Good luck.

Paul

Mar 12, 2015 | Office Equipment & Supplies

=>2x+3=0 ; taken x as common and 4-2=2 i.e. x(4-2) = 2x

=>2x=-3 ; +3's moved on the right-hand side, so it becomes -3

=>x=-3/2 ; 2 has been moved to the right-hand side and divided with -3 i.e 1/2 multiplied by -3 = -3/2

Good luck.

Thanks for using

Aug 13, 2010 | SoftMath Algebrator - Algebra Homework...

Use the rule for differentiating products of functions: ()' signifies derivative

(29*sin(2X)*sin(X))'= (29)'*sin(2X)*sin(X) +29* (sin(2X))'*sin(X) +29*sin(2X)*(sin(X))'

But

(29*sin(2X)*sin(X))'= 29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)

You could also have cast your formula in the form

sin(2X)*sin(X)= 1/2[ cos(2X-X)-cos(2X+X)]=1/2[cos(X)-cos(3X)]

then calculated the derivative of

29/2*[cos(X)-cos(3X)]

which is

29/2*[-si(X) +3*sin(3X)]

The challenge for you is to prove that the two forms are equivalent

29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)=29/2*[-si(X) +3*sin(3X)]

(29*sin(2X)*sin(X))'= (29)'*sin(2X)*sin(X) +29* (sin(2X))'*sin(X) +29*sin(2X)*(sin(X))'

But

- (29)'=0 derivative of a constant is zero
- (sin(2X))'=cos(2X)*(2X)'=2*cos(2X)
- (sin(X))'=cos(X)

(29*sin(2X)*sin(X))'= 29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)

You could also have cast your formula in the form

sin(2X)*sin(X)= 1/2[ cos(2X-X)-cos(2X+X)]=1/2[cos(X)-cos(3X)]

then calculated the derivative of

29/2*[cos(X)-cos(3X)]

which is

29/2*[-si(X) +3*sin(3X)]

The challenge for you is to prove that the two forms are equivalent

29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)=29/2*[-si(X) +3*sin(3X)]

Jun 21, 2010 | Vivendi Excel@ Mathematics Study Skills...

Hi,

Let's say your 3 numbers are a, b and c. Because they are consecutive and odd, you can write them like this:

a = 2x-1, b = 2x+1, c = 2x+3

Now, your problem looks like this:

2(2x-1)(2x+1) = (2x+1)(2x+3) + 7

2(4x^2-1) = 4x^2+6x+2x+3 + 7

8x^2-2 = 4x^2+8x+10

4x^2-8x-12 = 0

4(x^2-2x-3) = 0

x^2-2x-3 = 0

The solutions for this quadratic equation ( see more about this kind of equations here: http://en.wikipedia.org/wiki/Quadratic_equation ) are x=3 or x=-1. So, you've got two sets of solutions:

[1] a = 5, b = 7, c = 9

[2] a = -3, b = -1, c = 1

Take care,

Alex

Let's say your 3 numbers are a, b and c. Because they are consecutive and odd, you can write them like this:

a = 2x-1, b = 2x+1, c = 2x+3

Now, your problem looks like this:

2(2x-1)(2x+1) = (2x+1)(2x+3) + 7

2(4x^2-1) = 4x^2+6x+2x+3 + 7

8x^2-2 = 4x^2+8x+10

4x^2-8x-12 = 0

4(x^2-2x-3) = 0

x^2-2x-3 = 0

The solutions for this quadratic equation ( see more about this kind of equations here: http://en.wikipedia.org/wiki/Quadratic_equation ) are x=3 or x=-1. So, you've got two sets of solutions:

[1] a = 5, b = 7, c = 9

[2] a = -3, b = -1, c = 1

Take care,

Alex

Oct 15, 2009 | Google Computers & Internet

Edwin is 14. Ruel is 7.

In 3 years Edwin will be 17. Edwin will be 10. (3*10 - 13 = 17)

Let x=Ruel's age now, and 2x=Edwin's age now.

2x+3=3(x+3)-13

2x+3=3x+9-13

2x+3=3x-4

2x+7=3x

7=x

If you are going to post these, the least you could do is rate them as Fixya. :-) Thanks!

In 3 years Edwin will be 17. Edwin will be 10. (3*10 - 13 = 17)

Let x=Ruel's age now, and 2x=Edwin's age now.

2x+3=3(x+3)-13

2x+3=3x+9-13

2x+3=3x-4

2x+7=3x

7=x

If you are going to post these, the least you could do is rate them as Fixya. :-) Thanks!

Dec 11, 2008 | Computers & Internet

Paul 3 years younger to his friend peter. In seven years, the product of their ages would be 5 more than the product of their ages 5 years ago. How old are they now?

Oct 31, 2008 | Texas Instruments TI-30XA Calculator

B. 2((2-Y)/3)-Y=3

2(2-Y)-Y=9

4-2Y-Y=9

-3Y=9-4

-3Y=5

-Y=5/3

Y=-1.67

A. 3X+(-1.67)=2

3X-1.67=2

3X=2+1.67

3X=3.67

X=3.67/3

X=1.22

2(2-Y)-Y=9

4-2Y-Y=9

-3Y=9-4

-3Y=5

-Y=5/3

Y=-1.67

A. 3X+(-1.67)=2

3X-1.67=2

3X=2+1.67

3X=3.67

X=3.67/3

X=1.22

Nov 05, 2007 | SoftMath Algebrator - Algebra Homework...

247 people viewed this question

Usually answered in minutes!

×