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I understand, but at the same time, I don’t really understand.

You must be imagining things

You BUTTED in on Melody’s post!

thank you!!

ah I think I understand. Thank you!!

Well...if we wanted to  simplify  inches to feet we have

5 : 85    (simplify by dividing both by  5)

1 (inch) : 17 (feet)     ⇒   Answer 1

This must be imaginary understanding. Like imaginary numbers, it seems real but it’s not.

Thank you for answering! That's what I came up with also but the choices include -->

(1) 1 in. to 17 ft.

(2) 5 in. to 85 ft. 

(3) 2 in. to 35 ft.

(4) 3 in. to 51 ft. 

I'm a little confused. 

If an object that is 85 ft long is represented on a scale drawing as a line 5 in. long, what is the scale used to make the drawing?

85 ft  =   85 * 12  =  1020  inches

So the scale to the real  is :

5 : 1020      (simplify)

1 : 204 

Just divide one by the other!

1,092 / 42 =26 bushels per acre.

2(x+b) = ax+c    Simplify

2x + 2b   = ax  + c        If this has infinite solutions, then   a  = 2  and 2b  = c

So...note that

2x  + 2b  = 2x + c     will be true for any real x whenever  2b  =  c

If 3x-6y = 9z, which of the following expressions is equivalent to x^2 - 4xy + 4y^2

Dividing the first equation through by 3, we have

x - 2y  = 3z     (1)

Notice that   x^2 - 4xy  + 4y^2   factors as     (x -2y) ( x - 2y)        (2)

So subbing  (1)  into (2)  we get     (3z) (3z)  =  9z^2

\(\because \angle E = \angle E, \angle ABE = \angle CDE = 90º\therefore \text {By AA Similarity}, \triangle EBA \sim \triangle EDC.\\ \because \triangle EBA \sim \triangle EDC \therefore \frac{AB}{DC}=\frac{AE}{CE}.\\ \text{Filling in the numbers, we get}: \frac{4}{8}=\frac{5}{CE}.\\ \text{Solving for CE}: 4CE=5\cdot8\Rightarrow CE=10\)

The length of CE is 10. 

I hope this helped,

Gavin. 

\(\because \angle E = \angle E, \angle ABE = \angle CDE = 90º\therefore \text {By AA Similarity}, \triangle EBA \sim \triangle EDC.\\ \because \triangle EBA \sim \triangle EDC \therefore \frac{AB}{DC}=\frac{AE}{CE}.\\ \text{Filling in the numbers, we get}: \frac{4}{8}=\frac{5}{CE}.\\ \text{Solving for CE}: 4CE=5\cdot8\Rightarrow CE=10\)

The length of CE is 10. 

I hope this helped,

Gavin. 

Please no one else butt in because I actually want to help RB to understand.

Hi RB,

I like the way tyou said that.

I often say, I understand ... but ... I don't really understand.  

There  are different levels of understanding in mathematics.

Let's see if I can help.

I'll do the first one and then you can try the others.

1)  \(f(x)=2x-5\qquad find\;\;\quad f(6) \quad and \quad f(1)\)

For  f(6) you are replacing all the x'es with 6

so

\(f(6) = 2*6 - 5 = 12-5 = 7\)

-------------

For   f(1) you are replacing all the x'es with 1

so

\(f(6) = 2*1 - 5 = 2-5 = -3\)

NOW you have a go at at least one of the other questions. show me your working, and I will see if I can help you get to the point of  REALLY understanding       

Nvm got it again heh

Hint: Use similar triangles

Discovering a pattern  is often the key to  solving this type  of problem......

Let's look  at the  first 12  terms....we have

( 2001, 2002, 2003, 2000, 2005, 1998, 2007, 1996, 2009,  1994,  2011, 1992 .....  )

Note that the patern appears to  be that every  4th term decreases the series  by 4 

So we have that

2000  - 0(4)  =  2000  - 0   = 4th term

2000 - 1(4)  = 2000- 4 =  1996  =  8th term

2000 - 2(4)  = 2000 - 8 =  1992  = 12th term

So

2000 - 500(4) =2000 - 2000  = 0 =   2004th term 

2001, 2002, 2003, 2000, 2005, 1998, 2007, 1996, 2009, 1994, 2011, 1992, 2013, 1990, 2015, 1988, 2017............etc.
a_n = (-1)^n (-n + 2002 (-1)^n + 2) 
a(2004) = (-1)^2004(-2004 + 2002(-1)^2004 + 2)
a(2004) = 1(-2004 + 2002(1) + 2)
a(2004) = (-2004 + 2004)
a(2004) = 0

In triangle ABC, AB = 17, AC = 8, and BC = 15. Let D be the foot of the altitude from C to AB. Find the area of triangle ACD. 

Thank you, CPhill!

This is a right triangle  with AB forming the hypotenuse and AC and BC forming the legs.....the area will =  1/2 the product of the leg lengths...so we have

Area  = (1/2) (AC) ( BC)   =  (1/2)(8)(15)  =  (1/2) (120)  = 60  units^2

So.....letting ....AB  be the base of the triangle and CD  the altitude....then we have that

Area  = (1/2) (17) ( CD)

60  = (1/2) (17) (CD)

120/17 = CD

Now....ΔACB   is similar to  Δ ADC....this means that

AC / CB  = AD/ DC

8/15  = AD / (120/17)

(120/17)(8/15)  = AD

(120/15)(8/17)  = AD

(8)(8) / 17 = AD

64/17  = AD

Since  ADC is also a right triangle  with legs AD and  CD....its area  =  (1/2)(AD)(CD)  = (1/2)(64/17)(120/17)  =

(64 * 60) / 17^2  =

3840/289  units^2  ≈ 13.29 units^2

When x is divided by each of 4, 5, and 6, remainders of 3, 4, and 5 (respectively) are obtained.

What is the smallest possible positive integer value of x?

\(\begin{array}{|lrclcrcl|} \hline & x &\equiv& 3 \pmod 4 &\text{or} & x &\equiv& -1 \pmod 4 \\ & x &\equiv& 4 \pmod 5 &\text{or} & x &\equiv& -1 \pmod 5 \\ & x &\equiv& 5 \pmod 6 &\text{or} & x &\equiv& -1 \pmod 6 \\\\ \Rightarrow & x &\equiv& -1 \pmod{\text{lcm}(4,5,6)} \\ & x &\equiv& -1 \pmod{60} \\ & x &\equiv& -1+60 \pmod{60} \\ & \mathbf{x} & \mathbf{\equiv} & \mathbf{59 \pmod{60}} \\ \hline \end{array}\)

The smallest possible positive integer value x is 59

2

if g(z) is is less than f(z) then no matter what, adding any G(z) will not increase F(z)

thus what ever f(z) is, that is what f(z) + g(z) is

example

f(z) = x^2

G(z) must equal a function with x(degree 1) or 1(degree 0) or something 

x^2 + x +1 is still degree 2

doing some simplifing, we get

y> (-x+10)/2

without defining any limits to x or y

the answer is infinite because lets say x= anything over 10

then any y >0 would be positive and greater than our negative on the other side