All Answers 358093 Answers

please do this on a scientific calculator as i forgot one

Alternatively, you could use trail and error. So:

$${{\mathtt{1}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{2}}}^{{\mathtt{3}}} = {\mathtt{9}}$$

Too low

$${{\mathtt{3}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{4}}}^{{\mathtt{3}}} = {\mathtt{91}}$$

Still too low

$${{\mathtt{6}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{7}}}^{{\mathtt{3}}} = {\mathtt{559}}$$

Too high

$${{\mathtt{5}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{6}}}^{{\mathtt{3}}} = {\mathtt{341}}$$

Bingo

$${\mathtt{10}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{10}}{\mathtt{\,\times\,}}\left({{\mathtt{10}}}^{{\mathtt{10}}}\right)\right) = {\mathtt{110}}$$

Let n be the first number and (n + 1) be the second...so we have

n^3 + (n + 1)^3 = 341  expand

n^3 + n^3 + 3n^2 + 3n + 1 = 341   simplify

2n^3 + 3n^2 + 3n - 340 = 0

Using the Factor Theorem here might be a little difficult.....using the on-site solver is easier.....!!

$${\mathtt{2}}\left[{{n}}^{{\mathtt{3}}}\right]{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\left[{{n}}^{{\mathtt{2}}}\right]{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{n}{\mathtt{\,-\,}}{\mathtt{340}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{n}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{15}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{13}}\right)}{{\mathtt{4}}}}\\
{\mathtt{n}} = {\frac{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{15}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{13}}\right)}{{\mathtt{4}}}}\\
{\mathtt{n}} = {\mathtt{5}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{n}} = {\mathtt{\,-\,}}\left({\frac{{\mathtt{13}}}{{\mathtt{4}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4.841\: \!229\: \!182\: \!760\: \!208\: \!2}}{i}\right)\\
{\mathtt{n}} = {\mathtt{\,-\,}}{\frac{{\mathtt{13}}}{{\mathtt{4}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4.841\: \!229\: \!182\: \!760\: \!208\: \!2}}{i}\\
{\mathtt{n}} = {\mathtt{5}}\\
\end{array} \right\}$$

So n = 5 is the only "real" answer  ...and n+1 = 5+1 = 6

Check

5^3 + 6^3 =

125 + 216 =

341    ......√√√

First, we find the area of the base of the glasses, so using pi r^2, the area is $${\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{3.5}}}^{{\mathtt{2}}} = {\mathtt{38.484\: \!510\: \!006\: \!474\: \!967\: \!2}}$$cm2

next, find the area of the tray, so $${\mathtt{40}}{\mathtt{\,\times\,}}{\mathtt{25}}$$=1000cm2. Divide the area of the tray by the area of the glass to get your answer.

$${\frac{{\mathtt{1\,000}}}{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}\left({{\mathtt{3.5}}}^{{\mathtt{2}}}\right)\right)}} = {\mathtt{25.984\: \!480\: \!504\: \!799\: \!238\: \!5}}$$

so 25.9844805047992385 glasses or 25 whole glasses (we must round down in this case, otherwise, we go over the tray area)

-2+4--5=-2+4+5=7

Or we could rearrange this to something that looks more like something you have see before:

-2+4--5 -> 4--5-2 -> 4+5-2 =7

The question is a ambiguous so here are a number of answers.

If it is (2a^(3/5)-3b^(4/2))^2 then

2.297a^1.+-81b^4

if ((2a^3)/5-(3b^4)/2)^2 then

6052.84a^6+39601b^8

if ((2a^3)/5-(3b^(4/2)))^2 then

2.56a^6+81b^4

if ((2a^(3/5)-(3b^4)/2)^2 then

4a^1.2+1640.25b^8

Next time, use brackets around fractions so it is clear if it is a divide or fraction. Thats one of the problems of using   /   for both divide and fractions.

$$\left({\mathtt{14}}{\mathtt{\,\times\,}}{\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{100}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{18}}{\mathtt{\,\times\,}}{\mathtt{25}}{\mathtt{\,\times\,}}{\mathtt{100}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{22}}{\mathtt{\,\times\,}}{\mathtt{26}}{\mathtt{\,\times\,}}{\mathtt{100}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{100}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{100}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{11}}{\mathtt{\,\times\,}}{\mathtt{100}}\right) = {\mathtt{135\,900}}$$

Use the divide button. It will give you this -> /  which you can use for fractions.

Example:

23/48

127736+7567839=7695575

This is used when the angle is the unknown and the opposite and hypotenuse sides are known.

example

sin(x)=20/33

x=sin-1(20/33)

x=37.31 to 2dp

Some combinatorial problems in mathematics involve really large numbers (look up Graham's number for example), so it's possible these could be involved (though I can't think of any specific examples).

(1 + 3x)ⁿ = 1 + n*(3x) + n(n-1)*(3x)²/2! + n(n-1)(n-2)*(3x)³/3! + ...

Ratio of coefficients of x³ term to those of x² term:  n-2:1  so n-2 = 10.   n = 12.

Check

12(12-1)(12-2)*3³/3! = 5940

12(12-1)*3²/2! =  594

It depends upon the base....in a triangular pyramid (counting the base) there are 4. In a pyramid with a base of more than 3 sides, the number of triangles will just equal the number of sides of the base.

That would entirely depend of the type of pyramid. The best way is to look at the base of the pyramid and count the sides. That is the number of triangles the pyramid has. If the base is also a triangle, then add that triangle to your answer.

Using an additive identity (and assuming degrees), we have

sin 140 =

sin(90 + 50) =

sin90cos50 + sin50cos90 =

1*cos50 + p*0 =

cos50

Let p1 be the amount invested in the 7.5% interest account and p2 be that invested in the other.

Then 

p1 + p2 = 9099    ...(1)

0.075*p1 + 0.09*p2 = 699.84   ...(2)

Rewrite (1) as  p2 = 9099 - p1   ...(3)

Use (3) in (2)

0.075*p1 + 0.09*(9099 - p1) = 699.84

 -0.015*p1 + 0.09*9099 = 699.84

-0.015*p1 = 699.84 - 0.09*9099 

p1 =-( 699.84 - 0.09*9099)/0.015

$${\mathtt{p1}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{699.84}}{\mathtt{\,-\,}}{\mathtt{0.09}}{\mathtt{\,\times\,}}{\mathtt{9\,099}}\right)}{{\mathtt{0.015}}}} \Rightarrow {\mathtt{p1}} = {\mathtt{7\,938}}$$

p1 = $7938

Use this in (3)

p2 = 9099 - 7938

p2 = $1161

By combining like terms, we can simplify this to:

2a+b+c+3e+f+g+h+6i+2l+m+o+p+r+s+?+5(space)

2*6* 3√3*1/2 √3   we can re-write this as:

2*6*(1/2)*3*√3 *√3 =

12*(1/2) *3*3 =

6*9 =

54

Each number in the sequence is obtained by adding 5 on to the previous number.  Since the sequence starts with 2 this means the numbers alternately end in 2 and 7.  271 doesn't end in 2 or 7 and hence is not included in the sequence.

Expanding  (2 - px)^6 to the first three terms and  setting it equal to 64 + Ax + 135x^2, we have:

2^6 - (6)*2^5*(px) + 15*(2^4)*(px)^2 = 64 + Ax + 135x^2

64 - 192(px) + 240(px)^2 = 64 + Ax + 135x^2   ...   subtract 64 from both sides

-192(px) + 240(px)^2 = Ax + 135x^2

So, equating coefficients, we have

-192p = A      and  240p^2 = 135

So, solving for p^2 using the second equation, we have that p^2 = 135/240 = 9/16

And taking the positive square root, p = 3/4

And substituting for p in the first to find A, we have

-192(3/4) = A = -144

So...p = 3/4 and A = -144

Where???????

You can convert minutes to decimal degrees thusly:

Just divide the given minutes by 60. For instance, 30° 14′.....to convert the 14 minutes to decimal degrees, just divide 14 by 60 to get .2333....so we have 30.2333°........and that's it !!!!

The distance from Earth to Mars varies a lot, because Earth’s orbit around the sun is might smaller than Mars’ orbit. Earth is the third planet from the sun, and Mars is the fourth.

The minimum distance from the Earth to Mars is about 54.6 million kilometers. The farthest apart they can be is about 401 million km. The average distance is about 225 million km.

using 225 million km;

and the speed of light in vaccum 3*10^8 m/s(299 792 458 m/s^-1)

225000000/3*10^5(km /s) = 2250/35 = 64,285714285714285714285714285714 s