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why asinus ?

Hello ant101 :)

Solve for r

First you should note that  r cannot equal 2 or -1 becasue you cannot divide by 0.

\(\frac{r+3}{r-2}=\frac{r-1}{r+1}\\ (r+3)(r+1)=(r-1)(r-2)\\ r^2+4r+3=r^2-3r+2\\ 4r+3=-3r+2\\ 7r=-1\\ r=\frac{-1}{7}\)

check

(-1/7+3)/(-1/7-2) = -1.3333333333333333

(-1/7-1)/(-1/7+1) = -1.3333333333333333

Yep they are the same :)

Hello Mr Patel :)

1. There are five $5 and three $10 bills in a bag. A person reaches in and pulls out two bills. What is the probability of the sum of the bills will be $10, $15 or $20?

5,5,5,5,5,  10,10,10           8 notes altogether.

$10=$5+$5

\(P(5,5)=  \frac{5}{\not{8}^2}\times \frac{\not{4}^1}{7}=\frac{5}{14}\)

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$15    =   $5+$10    or     $10+$5

\(P(5,10)=  \frac{5}{8}\times \frac{3}{7}=\frac{15}{56}\\ P(10,5)=  \frac{3}{8}\times \frac{5}{7}=\frac{15}{56}\\ \text{P(sum of 15)}=\frac{30}{56}=\frac{15}{28} \)

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You can do the $20 yourself I think :)

150 = 10 times log x/10⁻¹²

I will assume that you mean

150 = 10 times log[ x/10⁻¹²]

\(150=10\times log(\frac{x}{10^{-12}})\\ 150=10\times log(x*10^{12})\\ 150=10\times[ log(x)+log(10^{12})]\\ 150=10\times[ log(x)+12log(10)]\\ 15=log(x)+12*1\\ 3=log(x)\\ 10^3=10^{log(x)}\\ 10^{log(x)}=10^3\\ x=1000 \)

or maybe you mean:

\(150 = 10 \times \frac{log (x)   }{   10^{-12}}\\ 15 = \frac{log (x)   }{   10^{-12}}\\ 15\times 10^{12} = log (x)  \\ x=10^{15\times 10^{12} }\\ x=10^{1.5\times 10^{13} }\\~\\ \text{WOW that is a big number !}\)

Okay then.... Good job on answering the riddle! 

Steps.

0.8 is a decimal expression already,so it's not clear what you are asking. If you want to write it as a fraction,it's 8/10 and if you want to write it as a percentage it's 80%.

 You are making me hungry

"Serena is wearing a pendant that was made by inscribing a square ruby in a sterling silver circle. The distance from the center of the pendant to its edge is 5.3 centimeters. Find the area of the pendant that is not covered by the ruby. Round to the nearest hundredth"

radius of circle: r = 5.3cm

Area of circular part of pendant;  A1 = pi*r² so A1 = pi*5.3² cm²

r is half the diagonal of the square ruby so the side length of the square is 2r/sqrt(2) → r*sqrt(2) 

Area of square ruby is:  A2 = (r*sqrt(2))² or A2 = 2*5.3² cm²

Area not covered by ruby:  A1 - A2 = (pi - 2)*5.3² cm² →    32.07 cm²

.

42.696

in an obstacle course race how many ways can five finalists be ordered?

5 choices for 1st, 4 choices for second etc

5*4*3*2*1=5!=120 ways

Find the equation of the line which contains the point (-2,3) and is parallel to the line 3x+5y=17

The travers are adding a new room to their house. The room will be a cube with volume 6859ft3. They are going to put in hardwood floors, and the contractor charges 10$ per square foot. How much will the hardwood floors cost?

\(\sqrt[3]{6859ft^3}=19ft\)

\( The \ hardwood \ floor \ costs \ (19ft)^2\times \frac{10$}{ft^2}=3610$\)

  !

the equation c^2 = p can have zero, one, or two solutions.

What can you say about the number of solutions of the equation c^3 = p?

The equation c ^ 3 = p may have one solution.

  !

Compute the definite integral:
 integral_0^1 ((1 - x)^4 x^4)/(x^2 + 1) dx
For the integrand ((1 - x)^4 x^4)/(x^2 + 1), cancel common terms in the numerator and denominator:
 = integral_0^1 ((x - 1)^4 x^4)/(x^2 + 1) dx
For the integrand ((x - 1)^4 x^4)/(x^2 + 1), do long division:
 = integral_0^1 (x^6 - 4 x^5 + 5 x^4 - 4 x^2 - 4/(x^2 + 1) + 4) dx
Integrate the sum term by term and factor out constants:
 = -4 integral_0^1 1/(x^2 + 1) dx + integral_0^1 x^6 dx - 4 integral_0^1 x^5 dx + 5 integral_0^1 x^4 dx - 4 integral_0^1 x^2 dx + 4 integral_0^1 1 dx
Apply the fundamental theorem of calculus.
The antiderivative of 1/(x^2 + 1) is tan^(-1)(x):
 = (-4 tan^(-1)(x)) right bracketing bar _0^1 + integral_0^1 x^6 dx - 4 integral_0^1 x^5 dx + 5 integral_0^1 x^4 dx - 4 integral_0^1 x^2 dx + 4 integral_0^1 1 dx
Evaluate the antiderivative at the limits and subtract.
 (-4 tan^(-1)(x)) right bracketing bar _0^1 = (-4 tan^(-1)(1)) - (-4 tan^(-1)(0)) = -π:
 = -π + integral_0^1 x^6 dx - 4 integral_0^1 x^5 dx + 5 integral_0^1 x^4 dx - 4 integral_0^1 x^2 dx + 4 integral_0^1 1 dx
Apply the fundamental theorem of calculus.
The antiderivative of x^6 is x^7/7:
 = -π + x^7/7 right bracketing bar _0^1 - 4 integral_0^1 x^5 dx + 5 integral_0^1 x^4 dx - 4 integral_0^1 x^2 dx + 4 integral_0^1 1 dx
Evaluate the antiderivative at the limits and subtract.
 x^7/7 right bracketing bar _0^1 = 1^7/7 - 0^7/7 = 1/7:
 = 1/7 - π - 4 integral_0^1 x^5 dx + 5 integral_0^1 x^4 dx - 4 integral_0^1 x^2 dx + 4 integral_0^1 1 dx
Apply the fundamental theorem of calculus.
The antiderivative of x^5 is x^6/6:
 = 1/7 - π + (-(2 x^6)/3) right bracketing bar _0^1 + 5 integral_0^1 x^4 dx - 4 integral_0^1 x^2 dx + 4 integral_0^1 1 dx
Evaluate the antiderivative at the limits and subtract.
 (-(2 x^6)/3) right bracketing bar _0^1 = (-(2 1^6)/3) - (-(2 0^6)/3) = -2/3:
 = -11/21 - π + 5 integral_0^1 x^4 dx - 4 integral_0^1 x^2 dx + 4 integral_0^1 1 dx
Apply the fundamental theorem of calculus.
The antiderivative of x^4 is x^5/5:
 = -11/21 - π + x^5 right bracketing bar _0^1 - 4 integral_0^1 x^2 dx + 4 integral_0^1 1 dx
Evaluate the antiderivative at the limits and subtract.
 x^5 right bracketing bar _0^1 = 1^5 - 0^5 = 1:
 = 10/21 - π - 4 integral_0^1 x^2 dx + 4 integral_0^1 1 dx
Apply the fundamental theorem of calculus.
The antiderivative of x^2 is x^3/3:
 = 10/21 - π + (-(4 x^3)/3) right bracketing bar _0^1 + 4 integral_0^1 1 dx
Evaluate the antiderivative at the limits and subtract.
 (-(4 x^3)/3) right bracketing bar _0^1 = (-(4 1^3)/3) - (-(4 0^3)/3) = -4/3:
 = -6/7 - π + 4 integral_0^1 1 dx
Apply the fundamental theorem of calculus.
The antiderivative of 1 is x:
 = -6/7 - π + 4 x right bracketing bar _0^1
Evaluate the antiderivative at the limits and subtract.
 4 x right bracketing bar _0^1 = 4 1 - 4 0 = 4:
Answer: |= 22/7 - π
 

The answer is just y=6x+90, cause you can't solve the equation unless you know the number for y

First, you need to know how much money it makes in a specific time like money per n second(s) or money per n second(s).

2.5 litres = 1,000 * 2.5 = 2,500 cubic cm

volume = π * radius² * height

  2,500 = π * r² * 20

\(\frac{2,500}{20\pi}=r^2 \\~\\ \sqrt{\frac{2,500}{20\pi}}=r \\~\\ 6.307 \approx r\)

Make the variable to its 3rd power

EX. X³ , The exponents means the same as dimensions in some places,

              like 23 ft³ is 3 dimension(Volume) and 23 ft² is 2 dimension(Area)

I think this is the question:

\(\frac{(xy)^3}{xy^3}=\frac{(xy)(xy)(xy)}{xyyy}=\frac{xxxyyy}{xyyy}=xx=x^2\)

But maybe you meant it how you actually have it written:

\(\frac{(xy)^3}{x}y^3=\frac{(xy)(xy)(xy)}{x}yyy=\frac{xxxyyyyyy}{x}=xxyyyyyy=x^2y^6\)

Haha that is pretty good!  I wouldn't have thought of potatoes...

I was thinking a blind man, but I thought that maybe too literal

'a' and 'b' is the legs and 'c' is the hypotenuse

The formula is a² + b² = c², which is the formula for Pythagorean Theorem

For Example, a=3 and b=4

                      a² + b² = c²                                                                           Pythagorean Theorem

                      3² + 4² = c²                                                                           Substitue 3 for a and 4 for b

                       9 + 16 = c²                                                                            Evaluate Powers

                                    25 = c²                                                                            Add

                                ^{\(\sqrt{25} = \sqrt{c^2}\)}                                                             Take square root of each side

                               5  = c                                                                Evaluate square root

                                                                                                       The hypotenuse for this equation is 5

potatos  :)

Appologies all around :)