All Answers 358087 Answers

2 times 5 is 10 and you minus 1 from 10 :)

cos(25)/21        should work fine

$${\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{25}}^\circ\right)}}{{\mathtt{21}}}} = {\mathtt{0.043\: \!157\: \!513\: \!668\: \!428\: \!6}}$$

24% of the people at a shopping mall are men. The number of boys is 3/8 of the number of men. The number of women is 4 times the number of boys. The number of girls is 504 more than the number of men. How many males are there in the shopping mall?

x are the number of all people.

m = men,  b = boys, w = women, g = girls

$$\small{
\begin{array}{rcl}
m&=& 0,24x\\\\
b&=& \dfrac{3}{8}m \\
w &=& 4\cdot b = 4\cdot \dfrac{3}{8}m = \dfrac{3}{2}m\\
g &=& 504 + m\\\\
x &=& m+b+w+g\\\\
x &=& m + \dfrac{3}{8}m+\dfrac{3}{2}m+504 + m\\\\
x &=& 2m+ \dfrac{3}{8}m+\dfrac{3}{2}m+504 \\\\
x &=& \dfrac{16+3+12}{8}m+504\\\\
x &=& \dfrac{31}{8}m+504\\\\
x &=& \dfrac{31}{8}\cdot 0.24x+504\\\\
x &=& 31\cdot 0.03x+504\\\\
x &=& 0.93x+504\\\\
x- 0.93x &=&504\\\\
0.07x &=&504\\\\
x &=& \dfrac{504\cdot 100}{7}\\\\
x &=& 7200\\\\
m &=& 0.24x\\
m &=& 0.24\cdot 7200 \\
\mathbf{m} & \mathbf{=} & \mathbf{1728}
\end{array}
}$$

There are 1728 males in the shopping mall

Boys = 648

Woman = 2592

Girls = 2232

Ok I think I have it now   

(2+i)x^2+(a-1)x+4-2i=0

$$\\(2+i)x^2+(a-1)x+4-2i=0\\\\
2x^2+(a-1)x+4+(x^2-2)i=0\\\\
2x^2+(a-1)x+4=0 \qquad AND \qquad x^2-2=0\\\\
2x^2+(a-1)x+4=0 \qquad AND \qquad x=\pm\sqrt2\\\\
$NOW consider$\\\\
2x^2+(a-1)x+4=0\\\\
2x^2+4=-(a-1)x\\\\$$

$$\\2*2+4=\pm\sqrt2(a-1)\\\\
8=\pm\sqrt2(a-1)\\\\
\frac{\pm 8}{\sqrt2}=a-1\\\\
\pm 4\sqrt2=a-1\\\\
a=1\pm 4\sqrt2\\\\$$

$$\\If \;\;x=+\sqrt2\qquad then\qquad a=1-4\sqrt2\\\\
If\;\; x=-\sqrt2\qquad then\qquad a=1+4\sqrt2\\\\$$

English books-35%, Tamil books-35*3/7=15%, Chinese books-(35+15)*3/10

Malay books-100-(35+15+15)=35%

(English and Malay books)=70% of total books=1190

→(Total books)=1700

Therefore, the number of tamil books is 1700*15/100=255

16% of the pens in a stationery shop are red. The number of black pens is 3/4 the number of red pens. The number of green pens is 5/7 the total number if red and black pens. The rest of the pens in the shop are blue. The number of blue pens is 180 more than the number of red, black and green pens. Find the total number of pens in the shop

x are the number of the pens.

$$\\\small{
\begin{array}{rcl}
\textcolor[rgb]{150,0,0}{\text{red}} &=& 0,16x\\\\
\text{black}& =& \dfrac{3}{4}\textcolor[rgb]{150,0,0}{\text{red}} \\\\
\textcolor[rgb]{0,150,0}{green} &=& \dfrac{5}{7}( \textcolor[rgb]{150,0,0}{\text{red}} +\text{black} ) = \dfrac{5}{7}\textcolor[rgb]{150,0,0}{\text{red}} + \dfrac{5}{7}\cdot \dfrac{3}{4}\textcolor[rgb]{150,0,0}{\text{red}} = \dfrac{5}{4} \textcolor[rgb]{150,0,0}{\text{red}} \\\\
\textcolor[rgb]{0,0,150}{blue}& =& 180 + ( \textcolor[rgb]{150,0,0}{\text{red}} +\text{black} + \textcolor[rgb]{0,150,0}{green} )
= 180 + \textcolor[rgb]{150,0,0}{\text{red}} +\dfrac{3}{4}\textcolor[rgb]{150,0,0}{\text{red}}+\dfrac{5}{4} \textcolor[rgb]{150,0,0}{\text{red}} = 180 + 3\cdot \textcolor[rgb]{150,0,0}{\text{red}} \\\\\\
x &=& \textcolor[rgb]{150,0,0}{\text{red}} +\text{Black}
+\textcolor[rgb]{0,150,0}{green}+\textcolor[rgb]{0,0,150}{blue}\\\\
x &=& \textcolor[rgb]{150,0,0}{\text{red}}
+\dfrac{3}{4}\textcolor[rgb]{150,0,0}{\text{red}}
+\dfrac{5}{4} \textcolor[rgb]{150,0,0}{\text{red}}
+180 + 3\cdot \textcolor[rgb]{150,0,0}{\text{red}} \\\\
x &=& 6\cdot \textcolor[rgb]{150,0,0}{\text{red}} +180\\\\
x &=& 6\cdot 0.16x+180\\\\
x &=& 0.96x+180\\\\
x-0.96x &=& 180\\\\
0.04x &=& 180\\\\
x &=& \dfrac{180\cdot 100}{4}\\\\
\mathbf{x}& \mathbf{=}& \mathbf{4500}
\end{array}
}$$

The total number of pens in the shop is 4500

red = 720

black = 540

green = 900

blue = 2340

This graph doesn't represent a function.

pens percentage=100- (Red pens+Black pens+Green pens) =100-48 =52 (52% of all pens)=(48% of all pens)+180 →(4% of all pens)=180 So the total number of pens in the shop is 4500 pens.

Blue pens percentage=100- (Red pens+Black pens+Green pens)

                              =100-48

                              =52

(52% of all pens)=(48% of all pens)+180

→(4% of all pens)=180

So the total number of pens in the shop is 4500 pens.

1,2,4,8,....2^(n-1).......2^29

a=1, r=2   n=30

$$\\S_n=\frac{a(r^n-1)}{r-1}\\\\
S_{30}=\frac{1(2^{30}-1)}{2-1}\\\\
S_{30}=2^{30}-1\\\\$$

$${{\mathtt{2}}}^{{\mathtt{30}}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{1\,073\,741\,823}}$$  cents

So the daily payment will add up to    $10,737,418.23

The up front amount is                        $2,000,000.00

Well the daily amount is more than 5 times the up front payment.

Thanks for your generosity jdh, but my answer is nonsense.  

9 out of 2=9/2=4.5

9 out of 0=0

9 out of 1=9/1=9

9 out of 5=9/5=1.8

I got

a=11/12 +ki

But i have not checked it and I would be extremely surprised if it is correct :)      

z^2=i  solve z

$$\small{\text{$
\begin{array}{rcl}
z^2&=&i \qquad | \qquad z^2= a+bi \quad a=0 $ and $ b = 1\\\\
r&=& \sqrt{a^2+b^2}=\sqrt{0^2+1^2}=\sqrt{1^2}=1\\\\
\varphi &=& \arctan{ \left(\dfrac{b}{a}\right) }=\dfrac{1}{0} \Rightarrow \varphi = \dfrac{\pi}{2}
\qquad
\varphi =
\begin{cases}
\arctan\left(\frac{b}{a}\right) & \text{for}\ a > 0,\ b \geq 0 \qquad \text{(1. Quadrant)}\\
\arctan\left(\frac{b}{a}\right) + 2\pi & \mathrm{f\ddot ur}\ a > 0,\ b < 0 \qquad \text{(4. Quadrant)}\\
\arctan\left(\frac{b}{a}\right) + \pi & \mathrm{f\ddot ur}\ a < 0 \qquad\qquad\quad \text{(2./3. Quadrant)}\\
\frac{\pi}{2} & \mathrm{f\ddot ur}\ a = 0,\ b > 0 \qquad \text{(positive Im-axis)}\\
\frac{3\pi}{2} & \mathrm{f\ddot ur}\ a = 0,\ b < 0 \qquad \text{(negative Im-axis)}\\
\text{indeterminate} & \mathrm{f\ddot ur}\ a = 0,\ b = 0 \qquad \text{(zero)}
\end{cases} \\\\
z^2&=& r\cdot e^{i\cdot \varphi}\\\\
z^2&=& 1\cdot e^{i\cdot \frac{\pi}{2}+2k\pi} \qquad | \qquad \sqrt{}\\\\
z_k &=& e^{i\cdot \frac{\pi}{2\cdot 2 }+\frac{2k\pi}{2} } \\\\
z_k &=& e^{i\cdot \frac{\pi}{4}+ k\pi } \\\\
k =0 \quad z_0 &=& e^{i\cdot \frac{\pi}{4}}\\\\
k =1 \quad z_1 &=& e^{i\cdot \frac{5\pi}{4}} \\\\
&& \boxed{~e^{i\varphi} =\cos{(\varphi)+i\sin{(\varphi)}} ~}\\\\
z_0 &=& e^{ i\frac{\pi}{4} } =\cos{(\frac{\pi}{4})+i\sin{(\frac{\pi}{4})}}\\\\
\mathbf{z_0} &\mathbf{=}& \mathbf{\dfrac{ \sqrt{2} }{2} +\dfrac{ \sqrt{2} }{2} i}\\\\
z_1 &=& e^{ i\frac{5\pi}{4} } =\cos{(\frac{5\pi}{4})+i\sin{(\frac{5\pi}{4})}}\\\\
\mathbf{z_1} &\mathbf{=}& \mathbf{-\dfrac{ \sqrt{2} }{2} -\dfrac{ \sqrt{2} }{2} i}
\end{array}
$}}$$

An equation must have both sides, however the equation that you wrote doesn't a right side.

Because of this, this equation cannot be solved.

eg) If something in real life weighs 2000t

2000/50=40

40/50=0.8

0.8/50=0.016t=16kg

6667/5678*888888=2963108148/2839

4 minus the absolute value of negative 6

→4-|-6|=4-6=-2

$${\mathtt{\,-\,}}{\frac{{\mathtt{25}}}{{\mathtt{4}}}} = \left({\frac{{\mathtt{5}}}{-{\mathtt{2}}}}\right){\mathtt{\,\times\,}}{\mathtt{x}} \Rightarrow {\mathtt{x}} = {\frac{{\mathtt{5}}}{{\mathtt{2}}}} \Rightarrow {\mathtt{x}} = {\mathtt{2.5}}$$

Hi Anshikasingh,

I don't remember seeing you before,  welcome to the forum :)

This is very easy

Take the base 2 number   10100111010110011111110000 

Split it into groups of 4 from the right hand side,

10  1001  1101  0110  0111  1111  0000 

in base 10 this would be a number between 0 and 15

In Hexidecimal it is a number between 0 and E

 $$10100111010110011111110000_2=$$  $$29D67E0_{16}$$

You would need to do more work to get the base 10 equivalent.

-25/4=-5/2x

-25/4*(-2/5)=x

x=5/2

2.3n+8=1.2n+2

1.1n=-6

n=-6/1.1=-5.4545..

C  =  π · d

9323  =  (22/7) · d

65261  =  22 · d

2966.41  =  d  (approximately)

9y - 4x  =  -30

Add 4x to both sides:

9y  =  4x - 30

Divide both sides by 9:

y  =  (4/9)x - (30/9)

y  =  (4/9)x - (10/3)