All Answers 356020 Answers

AP:     a, a+d,  a+2d,  a+3d,  .....

GP      \(\frac{1}{a},\; \frac{1}{a+d},\;\frac{1}{a+2d}, \dots\)

\(r=\frac{T_n}{T_{n-1}}=\frac{1}{a+d}\div\frac{1}{a}=\frac{1}{a+2d}\div\frac{1}{a+d}\;\;etc\\ \frac{a}{a+d}=\frac{a+d}{a+2d}\\ a(a+2d)=(a+d)(a+d)\\ a(a+d)+ad=a(a+d)+d(a+d)\\ ad=ad+d^2\\ d^2=0 \\ d=0\\ \)

Hence every term in the AP is 12 and every term in the GP is 1/12

Here is the full \(\LaTeX\) according to the order of operations:

1. Let \(f(x)= 3x+\frac{3}{x}+1\). Find the domain of \(f\)

2. Let \(q(x)=\frac{2}{\sqrt{x}}\). Find the domain of \(q\)

3.Let \(g(x)=3(x-6)^2+1\). Find the range of \(g\)

4.let \(f(a,b)= 2a-3b^2+7\). If \(f(k,3)=90\), then what is k?

5.Define the operation\( \heartsuit\) so that, \(a \heartsuit b=a+b/2\)

Find the value of \((5 \heartsuit 7) \heartsuit 11-5 \heartsuit (7 \heartsuit 11)\)

1.82 S.D.   corresponds to a Z-score of  .9656     (96.56 %)

women faster will be    (1 - .9656)  * 395   

a     +    2a      - b    =    3a - b

Here's a start...

In the equation 3a + 4b = 7, you can multiply through by 3 to get 9a + 12b = 21.

Next, in the equation 6a + 4b = k - 3a - 8b you can start by simplifying to 4a + 8b = k.

The slope formula is \({y_2 - y_1}\over{x_2-x_1}\). 

We can write this as an equation: \({{2a + 2 } \over {6+a} }= 7\)

Multiply both sides by \(6 + a\),

Then solve for a. 

in* not to

Or...

Solve the system of equations

a - b = 3 (1)

a + b = 5 (2)

Add (1) and (2) together and you'll get 

2a = 8 (3)

Now divide through by 2, then substitute a to the second equation (2), then find the ratio of a : b.

I do not understand the question.

Let x be Edwin's savings

2/5x be Ken's savings

2/5x + 20%(2/5x) + x - 55%x + $600

0.4x + 0.08x = 0.45x + $600

0.48x - 0.45x = $600

0.03x/0.03 = $600/0.03

       x = $20000

2/5 ($20,000) + 20% (2/5($20,00)) = $9600

I will use backward approach.

At last, all three of them had

Same amount of money.

So, All of them had $360/3 = $120

Before the end, Ken gave 2/3 of his money to Laura. So, 3/5 of Ken's money = $120

Ken's money in Previous Step = $200, Ken gave $80 to Laura. So, Laura's money in Previous Step = $(120 - 80) = $40

Step 2: - Tim gave 3/7 of his money to Ken.

So, 4/7 of Tim's money = $120

Tim's money in Step 2 = $210

Ken's money in Step 2 = $200 - $90 = $110 because Tim gave $90 to Ken.

Initially --> Laura gave half her money to Tim.

So, Laura's money initially = $40 * 2 = $80.

Tim's money initially = $(210 - 40) = $170

And, Ken's money initially = $110

Another way to get the same answer that BuilderBoi has:

Since the perimeter of the rhombus is 60, each side is 15.

Let's start by drawing the rhombus:

     each side is 15; 

     angle(C) = angle(A) = 129^o;

     angle(B) = angle(D) = 180^o - 129^o = 51^o.

To find the longer diagonal, DB, use the Law of Cosines on triangle(ABD):

    DB²  =  DA² + AB² - 2·DA·AB·cos(A)

    DB²  =  15² + 15² - 2·15·15·cos(129)

    DB²  =  733.194176

    DB    =  27.08

To find the shorter diagonal, AC, use the Law of Cosines on triangle(CDA):

     AC²  =  CD² + DA² - 2·CD·DA·cos(51)

Finish this to get the length of the shorter diagonal.

L==Laura,  T==Tim,   K==Ken
L - 1/2L ==1/2L - what L has left

T + 1/2L ==What T has now.
[T + 1/2L] x 3/7 ==3/7T + 3/14L - What T gave to K
K + 3/7T + 3/14L ==What K has now
4/7T + 4/14L ==What T has left...................................................(Equation 1)

[K + 3/7T + 3/14L] x 2/5==2/5K + 6/35T + 6/70L ==What K gave to L
2/5K + 6/35T + 6/70L + 1/2L ==What L has now........................(Equation 2)

3/5K + 9/35T + 9/70L ==What K has now....................................(Equation 3)

4/7T + 4/14L ==120................................(Equation 1)

[2/5K + 6/35T + 41/70L]==120...............(Equation 2)
[3/5K + 9/35T + 9/70L]==120.................(Equation 3)

Use, additions or subtractions or substitutions to get:
K ==110 - What Ken started with
L ==80 - What Laura started with
T==170 - What Tim started with

If you start on the x-axis at the left-end of the graph and draw vertical lines up (or down) into the graph and keep doing this until you get to the right-end of the graph and none of these vertical lines cut the graph in more than one point, it is a function.

If one or more of these vertical lines cut the graph in more than one point, it is not a function.

The exceptions occur when the denominator is zero:

     x² - 5x - 4  =  0     --->     (x - 4)(x - 1)  =  0     --->     x = 4  and  x = 1.

Can you complete the problem?

one-half of 75 is  ½ · 75  =  75/2  =  37½  =  37.5

Since f(x) is restricted to values  -1 < x < 1

and the function sqrt(x) further restricts this to the values 0 < x < 1,

the domain of j will be (0, 1).

I'd say   ....    You want as many places as possible so this is how it muxt be done.

11 111 111 111

Thanks!

You are very welcome MathyGoo13

Maths is my passion and I get great pleasure from teaching enthusiastic learners.   

So.. I appreciate you too :))

200 answers! That is impressive.  Congratulatons  BuilderBoi !

I have played with this for ages but I have not come up with anything helpful :((

The following graph should help

Multiply both sides by 5(a+b); distribute the results on both sides, collect the terms in a and b, then divide through by b.

Oh wow

Melody, I really really really really apperciate it!! 

And, you are a great teacher! I know this because I understand the question now :)

😊😊😊😊😊