All Answers 356458 Answers

Both 13 or 14 will work:

13^2 = 169 - All 4 digits different:1, 3, 6, 9

14^2 = 196 - All 4 digits different: 1, 4, 9, 6

The answer was 14/3 but thanks for helping

Using analytic geometry, you can work out that F = (8/4,6/4). Then the sum of the coordinates is 8/4 + 6/4 = 14/4 = 7/2.

$25 (2/3) (4/7)  = 25 (8/21) = $ 9.52             $  200/21       $  9  11/21  

EP: You have a typo at this stage:
7d - 12 + 3 =4d +12
7d - 4d =12 + 12 - 3
3d = 21
d = 7 - Daughter's age today
w = 37 - Mother's age today
3 Years from now, Mother will be =40 and Daughter will be = 10

(answer adapted from Balaji Reddy on Quora)

First of all, let's see how many one-row rectangles there are.

1 x 1 rectangles = (9–1)*(9–1) = 64

1 x 2 rectangles = (9–1)*(9–2) = 56

1 x 3 rectangles = (9–1)*(9–3) = 48

1 x 4 rectangles = 5 * 8 = 40

1 x 5 rectangles = 4 * 8 = 32

1 x 6 rectangles = 3 * 8 = 24

1 x 7 rectangles = 2 * 8 = 16

1 x 8 rectangles = 1 * 8 = 08

Total = 64+56+48+40+32+24+16+8 = 288 1-row rectangles!

There's a pattern here :D

Total one row rectangles = 8*1+8*2+…+8*8

= 8 * (1+2+…+8)

= 8 * (8*9/2) Since,n(n+1)/2 = sum of all numbers below n

=8 * 36

Similarly, Two row rectangles = 7 * (1+2+…+8)

=7 * 36

So all rows rectangles = 8*36+7*36+…+1*36

= 36 * (1+2+…+8)

= 36 * 36

It's \(\fbox{1296}\) :)

BTW, this also counts all the squares!

The ratio of 2: 1 for a mixture of 60 liters means:

2/3 x 60 = 40 liters of milk

1/3 x 60 = 20 liters of water

So, to  get the ratio down to 1 : 2, you have to add L liters of water:

M /[W + L] = 1 / 2

40 / [20 + L] = 1 / 2

80 =20 + L

L = 60 Liters of water have to be added to 60 Liters of mixture so that you have:

20 Liters of water + 60 Liters =80 Liters of water to 40 Liters of milk, or a ratio of 1 : 2.

d = daughter's age now

w = woman's age now        two years ago    w-2  = 7(d-2)     re-arrange to  w = 7d -12

and

w+3 = 4(d+3)        sub in the red equation

7d-12  +3  = 4d+12

3d = 21

d = 7      then mom = 37                *** edited ***     Thanx  , Guest for noticing my error.....I should have checked my answer !!

60 litres  2:1    means   1 out of   (2+1)     ,  or   1/3 is water = 20 liters       and 40 litres is milk

    40 litres of milk will be   1/3 of the new mixture       2/3 will be water = 80 litres    

                  so you will have to add 60 litrers of water to the original  20 liters

This should help:

Ex. (1,2,3,4,5,6,7,8,9,8,7,6,5,4,3,2, this is where the pattern restarts   1,2,3,4,5,6,7,8,9,8,7,6,5,4,3,2,      1...)

So there are 16 numbers before the pattern repeats.

so it will be  1000 mod 16

1000/16=62.5 = 62 remainder 8

the 8th number in the pattern is 8

so the 1000th digit is 8

y=x^6 + 3ax^5 + (3a^2 + 3) x^4 + (a^3 + 6a) x^3 + (3a^2 + 3) x^2 + 3ax + 1

Looks like about 1.77.

Go to the web2calc calcuator on the Home page and type in the following:

Then press enter.

I don't know what your teacher had in mind exactly, but here's one possibility:

A is at a radius of 3 units and an angle of 2*π/12 radians; 

B is at a radius of 4 units and an angle of 8*π/12 radians;

C is at a radius of 2 units and an angle of 21*π/12 radians.

I'll leave you to simplify the angles and express the results in the format requested.

What is the following value when expressed as a common fraction:
\(\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots + \frac{1}{2^{8}}+\frac{1}{2^{9}}+\frac{1}{2^{10}}\)?

\(\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots + \frac{1}{2^{8}}+\frac{1}{2^{9}}+\frac{1}{2^{10}}=\frac{2^9+2^8+2^7+...+2^2+2+1}{2^{10}} =\frac{512+256+128+...+4+2+1}{1024}=\frac{1023}{1024}\)

Your LaTex actually is displaying properly now but thanks for letting me know of your difficulties.

360/6= 60 degrees

So each parabola can take up 60 degrees from the origin.

Each half parabola takes up 30 degrees.

So the line y=(tan60) must be a tangent to the top parabola which is y=x^2+R

tan 60 = sqrt3

\(y=x^2+R\\ y'=2x\\ 2x=\sqrt3\\ x=\frac{\sqrt3}{2}\\ when \;\;x=\frac{\sqrt3}{2}\\ y=\sqrt3*\frac{\sqrt3}{2}=\frac{3}{2}\\ \text{The point of contact will be }( \frac{\sqrt3}{2},\frac{3}{2})\\~\\ y=x^2+R\\ \frac{3}{2}=( \frac{\sqrt3}{2})^2+R\\ \frac{3}{2}= \frac{3}{4}+R\\ R=\frac{3}{4}\)

Here are your parabolas

LaTex:

y=x^2+R\\
y'=2x\\
2x=\sqrt3\\
x=\frac{\sqrt3}{2}\\
when \;\;x=\frac{\sqrt3}{2}\\
y=\sqrt3*\frac{\sqrt3}{2}=\frac{3}{2}\\
\text{The point of contact will be }( \frac{\sqrt3}{2},\frac{3}{2})\\~\\
y=x^2+R\\
\frac{3}{2}=( \frac{\sqrt3}{2})^2+R\\
\frac{3}{2}= \frac{3}{4}+R\\
R=\frac{3}{4}

Subtract: \(\frac{x-2}{x+1}-\frac{3-12x}{2x^2-x-3}\)       [ \(2x^2-x-3=0\)   Assume: x = -1]

\((2x^2-x-3):(x+1)=2x-3\)

 \(\underline{2x^2+2x}\)

        \(-3x-3\)

        \(\underline{-3x-3}\)

                     \(0\)

 \(\frac{x-2}{x+1}-\frac{3-12x}{2x^2-x-3}=\) \(\frac{2x-3}{2x-3}\cdot \frac{x-2}{x+1}-\frac{3-12x}{2x^2-x-3}\)\(=\frac{2x^2-4x-3x+6-(3-12x)}{2x^2-x-3}\)     [expand with  (2x-3)]

\(=\frac{2x^2+5x+3}{2x^2-x-3}\)

  \((2x^2+5x+3):(x+1)=2x+3\)

   \(\underline{2x^2+2x}\)

               \(3x+3\)

               \(\underline{3x+3}\)

                         \(0\)

  \(\frac{2x^2+5x+3}{2x^2-x-3}\cdot\frac{x+1}{x+1}=\frac{2x+3}{2x-3}\)    [ shorten with (x+1)]

  \(\large \frac{x-2}{x+1}-\frac{3-12x}{2x^2-x-3}=\frac{2x+3}{2x-3}\)     \(x\in \mathbb R\ |\ \ x\notin \{-1.5\ ;-1\}\)

  !

I am not very comfortable with conics but ...

If m is positive then this has to be an ellipse.

\(\frac{x^2}{1}+\frac{y^2}{1/m}=4\\ \frac{x^2}{4}+\frac{y^2}{4/m}=1\\\)

a=2,  b=sqrt(4/m)

a is 2 (units from the centre),     

which is less than the focal distance of 4 so this must be the minor axis.

So it goes through (2,0) and (-2,0)

If c is the focal distance and a and b are the lengths of the major and minor axis  (i.e.   2 and  sqrt(4/m))

then

\(c^2=|a^2-b^2|\\ 16=\frac{4}{m}-4\\ 20=\frac{4}{m}\\ m=\frac{1}{5}=0.2 \)

\( x^2+0.2y^2=4 ​​ ​​\)

--------------------------------

If m is negative then it will be an hyperbola.

\(x^2+my^2=4 \)

\(focal \;\;distance = \sqrt{1+m^2}\\ 4= \sqrt{1+m^2}\\ 1+m^2=16 m^2=15\\ m=-\sqrt{15}\\ m\approx -3.873\)

\(\bf{\text{So I get  }\;\;\;m=0.2 \;\;\;or\;\;\; m=-\sqrt{15}}\)

LaTex:

c^2=|a^2-b^2|\\
16=\frac{4}{m}-4\\
20=\frac{4}{m}\\
m=\frac{1}{5}

focal \;\;distance = \sqrt{1+m^2}\\
4= \sqrt{1+m^2}\\
1+m^2=16
m^2=15\\
m=-\sqrt{15}

A right triangle is inscribed in a circle with a diameter 100 units long. What is the maximum area of the triangle, in square units?

The largest right-angled triangle inscribed in a circle is an isosceles right triangle.

area = 100² / 4

(1) Let WZYZ be a trapezoid with bases XY and Z. In this trapezoid,angle ZXW = 105,angle XWZ = 43 , and angle XYZ = 141. Find angle YXZ in degrees.

1/     Angle YXZ = 32 degrees.

(2) In trapezoid EFGH, line EF is parallel to line GH  and  P is the midpoint of side EH. If the area of triangle PEF is 8, and the area of triangle PGH  is 21, then find the area of trapezoid EFGH.

2/    The area of a trapezoid EFGH is 58 units squared.

( I used the right trapezoid with side lengths: EF = 8,  GH = 21, and EH = 4 )

(3) Quadrilateral ABCD  is a parallelogram. Let E  be a point on AB, and let F be the intersection of lines DE and BC. The area of triangle EBF is 1 and the area of triangle EAD  is 9 Find the area of parallelogram ABCD.

3/   Area of parallelogram ABCD = 24 u²

 (4) In trapezoid  ABCD,  AB is parallel to CD, and  AB < CD. The base CD  has a length of 8 The legs  AD and BC have lengths of 7 and the diagonal BD  has a length of 9 Find the area of the trapezoid.

4/    Area = 6√45

volume of original cylinder    -      volume of cone   =   volume remaining

pi r^2 h                                   -   1/3 pi r^2 h    = volume remaining

r  and h are the same for both of the figures

pi r^2 h ( 1 - 1/3)

pi  5²  6  (2/3)  = ..........   units³

Let the number be   xy        this is the same as  10x  +   y        you are given x+ y = 15      from this you can see that   y = 15-x

You are also given   yx = 30 + 6x    or    10y + x = 30 + 6x    sub in the red definition of y into this equation to get:

                                                                10(15-x)  + x = 30 + 6x

                                                                 150 - 10x  + x = 30 + 6x

                                                                   120 = 15x

                                                                     x = 8                      since  x+y = 15      then  y = 7

Original number is then   xy         87

   reversed number is    78    which   is equal to  30 + 6(8)