All Answers

#5

+2234

+3

Rosala, we may sometimes forget your birthday, but none of us will forget you!

This may be the post to which Melody is referring.

https://web2.0calc.com/questions/lets-end-the-day-by-learning-something#r4

This is why you are like springtime and flowers.

… ….

Your polite and pleasant manners and your willingness to communicate sincere, compassionate empathy, ranks you among the most unique of human beings.

I am sure, someday, no matter what you choose to do in life, you will be a diplomat – an ambassador to the world. Your posts represent a great beginning.

Happiness to you.

The ghost of web2.0calc

---

Here is the original reference:

https://web2.0calc.com/questions/which-sentence-contains-a-linking-verb-he-asked-questions-about-terms-he-did-not-understand-sarit-answered-all-of-the-questions-correctly#r13

Rosala is like springtime and flowers, even for parts of the world that are going into winter. ….

---

Here is another reference:

https://web2.0calc.com/questions/kitty-3-and-cphill-wait#r10

Rosala, do you know who the Ghost is\was?

GingerAleSep 18, 2017

#1

+130466

0

Note that

√[ 4 + √ [16 + 16a] ] can be written as

√[ 4 + 4√ [1 + a] ] =

2 √[ 1+ √ [1 + a] ]

Let this = 2x

So we have

2x + x = 6

3x = 6

x = 2

So...

√[ 1+ √ [1 + a] ] = 2 square both sides

1 + √ [1 + a] = 4 subtract 1 from both sides

√ [1 + a] = 3 square both sides again

1 + a = 9 → a = 8

CPhillSep 18, 2017

#1

+130466

0

1. 3y+4x=16

2x-y=18

Here's the graph of this one :https://www.desmos.com/calculator/ca7j4iemr4

2. y = -5x+6

10x+2y=12

Rearranging the first equation as 5x + y = 6

Multiplying by 2 on both sides

10x + 2y = 12 produces the same equation as the second one.....thus, these are the same line....so we will have infinite solutions

CPhillSep 17, 2017

#1

+130466

0

Let b₁ = a

So....b₃ = ar*2 and b₅ = ar*4

So

ar^2 - a = 16 → a [r^2 - 1] = 16 → a = 16 / [ r^2 - 1 ] (1)

ar^4 - ar^2 = 144→ a [ r^4 - r^2] = 144 → a = 144 / [ r^4 - r^2] (2)

Equating (1) and (2) we have that

16 / [ r^2 - 1 ] = 144 / [ r^4 - r^2] cross-multiply

16 [ r^4 - r^2] = 144 [ r^2 - 1 ] simplify

16r*2 [ r^2 - 1] = 144 [r^2- 1] divide through by 16

r^2 [ r^2 - 1] = 9 [r^2 - 1] divide through by [ r^2 - 1]

r^2 = 9 → r = 3 or -3

If r = ±3, then b₁ = a = 16 / [ 3^2 - 1] = 16 / 8 = 2

CPhillSep 17, 2017

#1

+2446

+2

The range is the set of all possible outputs that a function can produce.

$f(x)=4|x+5|$

The absolute value of any number will can result in a positive number or 0, and 4 multiplied by a positive number or 0 does not change this property at all. Therefore, the range is the following:

$\text{Range}:\{\mathbb{R}|\hspace{1mm}y\geq0\}$

This means that the range can be all nonnegative numbers. In interval notation, it would look like the following:

$\text{Range}:[0,+\infty)$

How do we find the intercepts? To do it without a graph, you can figure out by setting x=0 and y=0 and solving in each case. Let's do that.

$y=4\|x+5\|$	Plug in 0 for x. This time, we are solving for the y-intercept.
$y=4\|0+5\|$	Simplify inside the absolute value first.
$y=4\|5\|$
$y=4*5=20$	We have now determined the coordinates of the y-intercept.
$(0,20)$	This is the exact coordinates of the y-intercept.

Let's do the exact same process. This time, however, we set y=0 to find the x-intecept.

$y=4\|x+5\|$	Set y equal to 0.
$0=4\|x+5\|$	Divide by 4 on both sides.
$0=\|x+5\|$	The absolute value always splits an equation into its plus or minus. However, 0 is neither positive nor negative, so there aren't 2 equations that one can set up.
$x+5=0$
$x=-5$	We have now determined the x-intercept, as well.
$(-5,0)$

TheXSquaredFactorSep 17, 2017

#1

0

x = 0.8333r

10x = 8.333r

100x = 83.333r

100x - 10x = 83.333r - 8.333r = 75

90x = 75

18x = 15

6x = 5

x = 5/6

GuestSep 17, 2017

#1

+118703

+2

Express the following in the form a + bi, where a and b are real numbers: sqrt(i)

$\sqrt i=a+bi\\ i=(a+bi)^2\\ i=a^2+2abi-b^2\\ i=(a^2-b^2)+2abi\\ so\\ a^2-b^2=0\;\; and\;\ 2ab=1\\ (a-b)(a+b)=0\\ a=\pm b\\ but \;\;since\;\; 2ab=1, \;\;a=b\\ 2a^2=1\\ a=\pm\frac{1}{\sqrt{2}}\\ a=\pm\frac{\sqrt 2}{2}\\ so\\ \sqrt{i}=\pm(\frac{\sqrt 2}{2}+\frac{\sqrt 2i}{2})$

.

MelodySep 17, 2017

#1

0

11(5/6)

GuestSep 17, 2017

#2

0

thatnk you so much

GuestSep 17, 2017

#1

+130466

0

(n - 2) 180 / n = 176

Multiply both sides by n

(n - 2)180 = 176n

Simplify

180n - 360 = 176n

Add 360 to both sides, subtract176n from both sides

4n = 360

Divide both sides by 4

n = 90 [sides]

CPhillSep 17, 2017

#2

0

Let the time taken by your friend =t minutes, then you have:

3t =4(t -5), solve for t

t = 20 minutes taken by your friend

20 - 5 = 15 minutes taken by you.

And this makes sense since you can cover 1 miles in 15 minutes, while it takes her 20 minutes to cover 1 mile.

P.S. The reason that you don't need to divide by 60 is that the ratio of 3:4 remains the same, whether you divide or not.

GuestSep 17, 2017

#1

+130466

0

At 3 mph....your friend covers 3 / 60 miles per minute

At 4 mph......you cover 4/ 60 miles per minute

So....every minute....you close [ 4 - 3 ]/ 60 = 1/ 60 miles on your friend

And when she has been walking for 5 minutes......she will be 5 * [3 / 60 ] = 15/60 = 1/4 of a mile ahead of you when you start

So......the number of minutes, m, it will take to catch your friend can be found by :

( 1/60) m = 1/4 multiply both sides by 60

m = 60 / 4 = 15 minutes

CPhillSep 17, 2017

#4

-4

$10,000 x [12%] / 2 =$10,000 x 0.06 =$600 interest earned for the 1st 6 months.

$10,000 + $600 =$10,600 principal + interest for the 1st 6 months.

$11,130 / $10,600 =1.05

[1.05 -1] x 100 =5% This is the effective annual interest on the 2nd certificate. Or you can:

1.05^1/2 =[1.0246950 - 1] x 100=2.4695% semi-annual copound rate. Or:

2.4695 x 2 =4.939% - This is called nominal annual rate compounded semi-annually, which is equivalent to 5% effective annual rate.

GuestSep 17, 2017

#1

+2446

+5

A constant term is the term that is unchanging. 6, for example, is a constant. 12.5 is a constant; they don't change.

To figure out $f(g(x))$ , let's break this down bit by bit. Since we already know that $g(x)=x^3+5x^2+9x-2$ , this means that $f(g(x))=f(x^3+5x^2+9x-2)$ .

$f(x)=x^3-6x^2+3x-4$	Now, substitute f(g(x)) into all instances of x.
$f(g(x))=(x^3+5x^2+9x-2)^3-6(x^3+5x^2+9x-2)^2+3(x^3+5x^2+9x-2)-4$	Luckily, however, we only care about the constant terms. Let's deal with one term at a time.
$(x^3+5x^2+9x-2)^3$	We only care about the constant term, so do -2^3=-8
$(-2)^3=-8$	Let's worry about the second term.
$-6(x^3+5x^2+9x-2)^2$	Let's do the exact same process.
$-6(-2)^2=-64=-24$	And of course, the next term, as well.
$3(x^3+5x^2+9x-2)$
$3*-2=-6$	And the final term, which happens to be a constant.
$-4=-4$	Now, add all of those together to get the constant term.
$-8-24-6-4 = -42$	This is value of the constant term.