All Answers 356020 Answers

The product is of course: 2 x 1 x 2 =4

Using Continuous Fractions Method, we have:

1/.375 =2 2/3 =2

1/(2/3)= 1 1/2 =1

1/(1/2) =2, Therefore:

a=2     b=1     c=2

Like this:\(\begin{align} -\frac{x}{2}-4&=-7 \; /\cdot(-2)\\ \frac{x(-2)}{-2}-4(-2)&=-7(-2)\\ x+8&=14\\ x&=14-8\\ x&=6 \end{align}\)

f a + b = 1 and a^2 + b^2 = 2, what is the value of a^3 + b^3? Express your answer as a common fraction.

If a + b  = 1....square both sides

a^2 + 2ab + b^2  = 1

(a^2 + b^2)  + 2ab  = 1

2 + 2ab = 1

2ab  =  1- 2

2ab  = -1

ab  = -1/2  →  -ab  = 1/2

So

a^3 + b^3  =

(a + b) (a^2 - ab + b^2) =

(a + b) ( [a^2 + b^2]  - ab)  =

(1) (  2 + 1/2) =

(1)( 5/2)  =

5 / 2

(4x-x)5 = 

(3x)*5 =

15x

y= [6x-2] / [2x+2]

The domain will include all x values except x = -1........this would make the denominator 0

The funcrion will have a horizontal asymptote at y = 6/2   = 3

So.......the range will be  all y except y = 3

Here's a graph :  https://www.desmos.com/calculator/2kossvoaj7

907632 mod 33 =0

its 2500

How do you simplify 2log4 9 - log3 ​3?

umm.. im not really good at english but i hope i can help you.

this is a sequence of triangular numbers.

I'll give you a general term :

n(n+1)

    2

n is the number of numbers you have in your sequence, like:

1+2+3+4+5+6+7

then you'll put it like this:

7(7+1)

   2

hope you'll understand xdddddddd

That depends on what information you have to begin with.

The following relations apply to circles:

\(area = \pi * r^2\)

\(circumference = 2*\pi*r\)

In which the Greek letter pi is simply a number, approximately 22/7 but in reality 3,1415... endless string of numbers.

The meaning of this number is as such. If you have a radius then there is a specific angle at which the length of the described arc is equal to the radius. This angle is defined as one radian. The number of radians that fit in one circle, or 360 degrees, is exactly 2 pi radians. Therefore the circumference becomes 2 * pi * r.

If you know the radius of the circle, or the diameter, which is twice the radius (note D = 2*r) you can easily compute the circumference.

If you have the area of a circle, you can divide by pi and take the square root:

\(r=\sqrt{area\over\pi}\)

From which you can then calculate the circumference.

Standard deviation is a measure of a set of data, measuring how close the data group around their central value (the mean).

It does not apply to a single number (18912/11).

0,593

2sec(x+60)=5sec(x-20)

\(\begin{array}{|rcll|} \hline 2\cdot \sec(x+60^{\circ}) &=& 5\cdot \sec(x-20^{\circ}) \\ \text{We substitute } x = y-20^{\circ} \\\\ 2\cdot \sec(y-20^{\circ}+60^{\circ}) &=& 5\cdot \sec(y-20^{\circ}-20^{\circ}) \\ 2\cdot \sec(y+40^{\circ}) &=& 5\cdot \sec(y-40^{\circ}) \quad | \quad \sec(\phi) = \frac{1}{\cos(\phi)}\\ \frac{ 2 } { \cos(y+40^{\circ}) } &=& \frac{ 5 } { \cos(y-40^{\circ}) } \\ \frac{ \cos(y-40^{\circ}) } { \cos(y+40^{\circ}) } &=& \frac{ 5 } { 2 } \quad | \quad \cos(y-40^{\circ}) = \cos(y)\cos(40^{\circ})+\sin(y)\sin(40^{\circ}) \\ \frac{ \cos(y)\cos(40^{\circ})+\sin(y)\sin(40^{\circ}) } { \cos(y+40^{\circ}) } &=& \frac{ 5 } { 2 } \quad | \quad \cos(y+40^{\circ}) = \cos(y)\cos(40^{\circ})-\sin(y)\sin(40^{\circ}) \\ \frac{ \frac{\cos(y)\cos(40^{\circ})+\sin(y)\sin(40^{\circ})} {\cos(y)\cos(40^{\circ})} } { \frac{\cos(y)\cos(40^{\circ})-\sin(y)\sin(40^{\circ})} {\cos(y)\cos(40^{\circ})} } &=& \frac{ 5 } { 2 } \\ \frac{ 1+\tan(y)\tan(40^{\circ}) } {1-\tan(y)\tan(40^{\circ}) } &=& \frac{ 5 } { 2 } \\\\ 2\cdot [~ 1+\tan(y)\tan(40^{\circ}) ~] &=& 5\cdot [~ 1-\tan(y)\tan(40^{\circ}) ~] \\ 2 + 2\cdot \tan(y)\tan(40^{\circ}) &=& 5 - 5 \cdot \tan(y)\tan(40^{\circ}) \\ 7\cdot \tan(y)\tan(40^{\circ}) &=& 3 \\ \tan(y) &=& \frac37\cdot \frac{1}{\tan(40^{\circ})} \quad | \quad y=x+20^{\circ}\\ \tan(x+20^{\circ}) &=& \frac37\cdot \frac{1}{\tan(40^{\circ})} \\ x+20^{\circ} &=& \arctan{\left( \frac37\cdot \frac{1}{\tan(40^{\circ})} \right)} \pm n\cdot 180^{\circ}\\ x &=& -20^{\circ} + \arctan{\left( \frac37\cdot \frac{1}{\tan(40^{\circ})} \right)} \pm n\cdot 180^{\circ}\\ x &=& -20^{\circ} + \arctan{\left( 0.51075153968 \right)} \pm n\cdot 180^{\circ}\\ x &=& -20^{\circ} + 27.0557431286^{\circ} \pm n\cdot 180^{\circ}\\\\ \mathbf{x} &\mathbf{=}& \mathbf{7.0557431286^{\circ} \pm n\cdot 180^{\circ} } \qquad n \in N\\ \hline \end{array}\)

you now what is 9 + 10 =?

If log_(8)3=P and log_(3)5=Q, express log_(10)5 in terms of P and Q.

Your answer should no longer include any logarithms.

\(\begin{array}{|rcll|} \hline \log _b \left( x \right) &=& \dfrac{ \log _c \left( x \right) } { \log _c \left( b \right) } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline &P &=& \log_8 \left( 3 \right) \\\\ &P &=& \dfrac{ \log _{10} \left( 3 \right) } { \log _{10} \left( 8 \right) } \\\\ &P &=& \dfrac{ \log _{10} \left( 3 \right) } { \log _{10} \left( 2^3 \right) } \\\\ &P &=& \dfrac{ \log _{10} \left( 3 \right) } { 3\cdot \log _{10} \left( 2 \right) } \\ \hline (1) & \log _{10} \left( 2 \right) &=& \dfrac{ \log _{10} \left( 3 \right) } {3P} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline &Q &=& \log_3 \left( 5 \right) \\\\ &Q &=& \dfrac{ \log _{10} \left( 5 \right) } { \log _{10} \left( 3 \right) } \\\\ \hline (2) & \log _{10} \left( 3 \right) &=& \dfrac{ \log _{10} \left( 5 \right) } {Q} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \log _{10} \left( 5 \right) &=& \log _{10} \left( \frac{10}{2} \right) \\\\ \log _{10} \left( 5 \right) &=& \log _{10} \left( 10 \right)- \log _{10} \left( 2 \right) \quad &| \quad \log _{10} \left( 10 \right) = 1\\\\ \log _{10} \left( 5 \right) &=& 1- \log _{10} \left( 2 \right) \quad &| \quad \log _{10} \left( 2 \right) = \dfrac{ \log _{10} \left( 3 \right) } {3P}\\\\ \log _{10} \left( 5 \right) &=& 1- \dfrac{ \log _{10} \left( 3 \right) } {3P} \quad &| \quad \log _{10} \left( 3 \right) = \dfrac{ \log _{10} \left( 5 \right) } {Q}\\\\ \log _{10} \left( 5 \right) &=& 1- \dfrac{ \dfrac{ \log _{10} \left( 5 \right) } {Q} } {3P} \\\\ \log _{10} \left( 5 \right) &=& 1- \dfrac{ \log _{10} \left( 5 \right) } {3PQ} \\\\ \log _{10} \left( 5 \right) + \dfrac{ \log _{10} \left( 5 \right) } {3PQ} &=& 1 \\\\ \log _{10} \left( 5 \right)\cdot \left( 1+ \dfrac{ 1 } {3PQ} \right ) &=& 1 \\\\ \log _{10} \left( 5 \right)\cdot \left( \dfrac{ 3PQ+1 } {3PQ} \right ) &=& 1 \\\\ \mathbf{\log _{10} \left( 5 \right)} &\mathbf{=}& \mathbf{\dfrac{3PQ}{ 3PQ+1 } }\\ \hline \end{array}\)

Geno, since when is 115 + 115 + 230 + 345 = 595? 

35th term of Arithmetic sequence 3,7,11,15,19

\(\small{\text{We have $t_x=t_1=3$ and $t_y=t_{2}=7$ and we want $t_z=t_{35}=?$ }}\\\\ \small{\text{$\boxed{~~t_z = t_x\cdot \left( \dfrac{y-z}{y-x}\right) + t_y\cdot \left(\dfrac{z-x}{y-x}\right) ~~} $}}\\\\ \small{\text{$ \begin{array}{rcl} t_{35} &=& t_1\cdot \left( \dfrac{2-35}{2-1}\right) + t_{2}\cdot \left(\dfrac{35-1}{2-1}\right) \\\\ t_{35} &=& 3\cdot \left( \dfrac{-33}{1}\right) + 7\cdot \left(\dfrac{34}{1}\right) \\\\ t_{35} &=& -3\cdot 33 + 7\cdot 34 \\ t_{35} &=& -99 + 238 \\ \mathbf{t_{35}} & \mathbf{=} & \mathbf{139}\\ \hline \\ \end{array} $}}\)

Hello, good Morning Calpal!

x² + x +1 > 0

Solve the inequality

x ∈ {ℝ}

Greting asinus :-   !

Find the remainder when $1^3 + 2^3 + 3^3 + \dots + 100^3$ is divided by 6.

\(\text{The sum of } ~ 1^3 + 2^3 + 3^3 + \dots + n^3 = \left[ \frac{ n*(n+1) } {2} \right]^2\\ \text{The sum of } ~ 1^3 + 2^3 + 3^3 + \dots + 100^3 = \left[ \frac{ 100*(100+1) } {2} \right]^2 \)

\(\begin{array}{|rcll|} \hline && 1^3 + 2^3 + 3^3 + \dots + 100^3 \pmod 6 \\ &\equiv & \left[ \frac{ 100*(100+1) } {2} \right]^2 \pmod 6 \\ &\equiv & \left[ \frac{ 100*101 } {2} \right]^2 \pmod 6 \\ &\equiv & \left[ 50*101 \right]^2 \pmod 6 \\ &\equiv & 5050^2 \pmod 6 \qquad &| \qquad 5050 \pmod 6 = 4 \\ &\equiv & 4^2 \pmod 6 \\ &\equiv & 16 \pmod 6 \\ &\equiv & 4 \pmod 6 \\ \hline \end{array}\)

What is the remainder when $13^{51}$ is divided by 5?

\(\begin{array}{|rcll|} \hline && 13^{51} \pmod {5} \qquad &| \qquad 13 \pmod 5 \equiv 3 \\ &\equiv& 3^{51} \pmod {5} \qquad &| \qquad 3^4 \equiv 1 \pmod 5 \\ &\equiv & 3^{4\cdot 12+3} \pmod {5} \\ &\equiv & 3^{4\cdot 12} \cdot 3^3 \pmod {5} \\ &\equiv & (3^{4})^{12} \cdot 3^3 \pmod {5} \qquad &| \qquad 3^4 \equiv 1 \pmod 5 \\ &\equiv & (1)^{12} \cdot 3^3 \pmod {5} \qquad &| \qquad (1)^{12}=1\\ &\equiv & 1 \cdot 3^3 \pmod {5} \\ &\equiv & 27 \pmod {5}\\ &\equiv & 2 \pmod {5}\\ \hline \end{array}\)

What is the remainder when $2^{2005}$ is divided by 7?

\(\begin{array}{|rcll|} \hline && 2^{2005} \pmod {7} \qquad &| \qquad 2^6 \equiv 1 \pmod 7 \\ &\equiv & 2^{6\cdot 334+1} \pmod {7} \\ &\equiv & 2^{6\cdot 334} \cdot 2 \pmod {7} \\ &\equiv & (2^{6})^{334} \cdot 2 \pmod {7} \qquad &| \qquad 2^6 \equiv 1 \pmod 7 \\ &\equiv & (1)^{334} \cdot 2 \pmod {7} \qquad &| \qquad (1)^{334}=1\\ &\equiv & 1 \cdot 2 \pmod {7}\\ &\equiv & 2 \pmod {7}\\ \hline \end{array} \)

Solve for x:
x^2-6 = 3

Add 6 to both sides:
x^2 = 9

Take the square root of both sides:
Answer: |x = 3 or x = -3

is 890,000 square miles larger than 460,000 miles?

You CANNOT compare " square miles" to "straight miles". If you wish to compare them, you must either take square root of "square miles" or square the "straight miles". So, we have:

Sqrt(890,000) =943.40 This is the sraight miles on one side of the square. So, it doesn't compare to 460,000 straight miles.

Expand the following:
(3 x+2)^4 = 232

(3 x+2)^4 = sum_(k=0)^4 binomial(4, k) 2^(4-k) (3 x)^k = binomial(4, 0) (2)^4+binomial(4, 1) (2)^3 (3 x)+binomial(4, 2) (2)^2 (3 x)^2+binomial(4, 3) (2) (3 x)^3+binomial(4, 4) (3 x)^4 = 81 x^4+216 x^3+216 x^2+96 x+16:
81 x^4+216 x^3+216 x^2+96 x+16 = 232

Subtract 232 from both sides of 81 x^4+216 x^3+216 x^2+96 x+16 = 232:
81 x^4+216 x^3+216 x^2+96 x-232+16 = 232-232

232-232 = 0:
81 x^4+216 x^3+216 x^2+96 x-232+16 = 0

Grouping like terms, 81 x^4+216 x^3+216 x^2+96 x-232+16 = 81 x^4+216 x^3+216 x^2+96 x+(16-232):
81 x^4+216 x^3+216 x^2+96 x+(16-232) = 0

16-232 = -216:
Answer: |81 x^4+216 x^3+216 x^2+96 x+-216 = 0