CPhill

I think Christina just wanted somone to verify this....you have done so, rosala !!!

Using the first equation, we have 

y = 5 - 5x        And substituting this into the second equation, we have

4x - 7(5 - 5x) = 43

4x - 35 + 35x = 43     simplifying, we have

39x - 35 = 43            add 35 to both sides

39x = 78                   divide by 39 on both sides

x = 2    

And, using y = 5 - 5x, we have    

y = 5 - 5(2)  

y = 5 - 10

y = -5

You should verify that these answers are correct......!!!

I think the questioner wants to know this:

The sum of the "x" components of a 270°  vector with a length of 4 and a 60° vector with a length of 4.

So, for the 270° vector we have, < 0. 4 >  and for the 60° vector we have < 2 ,  2√3 > .

So, as Melody has said, the sum of the "x" components is just  2.

Let's take the "arctan" part first....remember, this will return some angle

arctan(12/5) = 67.38013505196°

So.....2*arctan(12/5) = 2* 67.38013505196°   = 134.76027010392°

And...... sin( 134.76027010392°) = 0.710059171598

This is modeled by.....

h(t) =  ax^2 + bx  

and

h(1) =  a(1)^2 + b(1) =  110  →     110 = a + b    (1)

and

h(2) =  a(2)^2 + b(2) =  200  →     200 = 4a + 2b   (2)

Multiplying (1) by -2 and adding it to (2) gives us

-20 = 2a       so   a = -10

And substituting this back into (1) we can find b

110= -10 + b       so b = 120

So our function is

h(t) = -10x^2 + 120x

So....after 4 seconds

h(4) =   -10(4)^2 + 120(4)  =  320 ft.

OK....

f(x) = 5^-x   can be wrtten as  1/5^x

Note that we can put any real number in for "x" because  an exponential function never = 0 , so the denominator never = 0   So "domain" relates to x....and the domain is just (-∞ , ∞).....just as you suspected........!!!

The range - which relates to "y" - is trickier....as x is more and more negative, the function grows larger and as x gets more positive, the function gets very close to 0 (but not actually 0).  So the range is (0 , ∞). Here's where a graph may help!!

Note how the graph gets "close" to 0 on the right, but is unbounded on the left!!

Hope this helps !!!

Graphing is sometimes easier, but we can often analyze the function, too.......

Sometimes....I use both....just to verify my answer

Do you have a particular function in mind??

OK.....I guess all of us can put ourselves on pedestals   .....!!!!

After all....maybe we can't ALL be smart eggs.....but at least we're "colorful"......!!!!!!

justin is using 120 inches of wire to biuld two structures one rectangle that is five times as long as it it is wide and one square whose side lenght is the same as the width of the rectangle. what will be the exact area of his square?

So ......the perimeter of the rectangle is given by 2(L + W)= 2(5W + W) = 2(6W) = 12W

And the perimeter of the square is given by 4W

So

12W + 4W = 120

16W = 120

W = 7.5 inches

So...the area of the square is just   (7.5)^2 =  56.25 in²

Find the total cost of producing 5 widgets. Widget Wonders produces widgets. They have found that the cost, c(x), of making x widgets is a quadratic function in terms of x. The company also discovered that it costs $15.50 to produce 3 widgets, $23.50 to produce 7 widgets, and $56 to produce 12 widgets.

OK...so we have

a(7)^2 + b(7) + c =  23.50 →   49a + 7b + c = 23.50    (1)

a(3)^2 + b(3) + c =  15.50 →     9a  + 3b + c = 15.50         subtracting the second equation from the first, we have

40a + 4b = 8  → 10a + b = 2    (2)

Also

a(12)^2 + b(12) + c = 56  → 144a + 12b + c = 56   and subtracting (1) from this gives us

95a + 5b = 32.50

And using(2)  we have

95a + 5b = 32.50   (3)

10a + b  =    2.00        multiplying the second equation by -5  and adding  this to (3) ,we have

45a = 22.50      divide both sides by 45    and a = 1/2   and using (2) to find b, we have

10(1/2) + b = 2

5 + b = 2      b = -3

And we can use 9a  + 3b + c = 15.50 to find "c"

9(1/2) + 3(-3) + c = 15.50

9/2 - 9  + c = 15.50

-4.5 + c = 15.50      

c = 20

So our function is

c(x) = (1/2)x^2 - (3)x  + 20

And the cost to produce 5 widgets is = $17.50

Here's the graph of our function