All Answers 358504 Answers

$${\frac{{\mathtt{7}}}{{\mathtt{8}}}} = {\mathtt{0.875}}$$

Well the vertex is at P but whether is wants the reflex angle or the other one is not clear.

I'd go for the smaller angle, most times that would be what is wanted. 

I think the sine rule might be the simplest approach here:

h/sin(H) = j/sin(J)

$${\mathtt{h}} = {\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{132}}^\circ\right)}{\mathtt{\,\times\,}}{\mathtt{31}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{21}}^\circ\right)}}} \Rightarrow {\mathtt{h}} = {\mathtt{64.284\: \!458\: \!526\: \!596\: \!557\: \!1}}$$

or h = 64 to the nearest whole number

I've assumed h is opposite angle H, j is opposite angle J etc.

.

x-2y=-2       x^2-y^2=7

$$x-2y=-2 \qquad x=2y-2 \\\\
x^2-y^2=7 \qquad (2y-2)^2-y^2=7\\
4y^2-8y+4-y^2=7\\
3y^2-8y+4=7\\
3y^2-8y-3=0\\\\
\small{\text{
$
y_{1,2}=\dfrac{8\pm\sqrt{64-4*3*(-3)} } {2*3}
=\dfrac{8\pm\sqrt{64-36} } {6}
=\dfrac{8\pm\sqrt{100} } {6}
=\dfrac{8\pm10 } {6}
$
}}\\\\
\small{\text{
$
y_1=\dfrac{8+10 } {6}=\dfrac{18 } {6}=3
$
}}\\\\
\small{\text{
$
y_2=\dfrac{8-10}{6}=-\dfrac{2}{6}=-\dfrac{1}{3}
$
}}\\\\
\small{\text{
$
x_1=2*3-2 = 4
$
}}\\
\small{\text{
$
x_2=2*(-\frac{1}{3})-2 = -2\frac{2}{3}
$
}}$$

(m+3)2-(m+3)(m-3)

$$(m+3)2-(m+3)(m-3)\\
(m+3)(2-(m-3))\\
(m+3)(2-m+3)\\
(m+3)(5-m)\\$$

solve the given equation for the variable:

6x - 3(x - 1) = 4(2x + 1)

$$6x - 3(x - 1) = 4(2x + 1)\\
6x-3x-3(-1) = 4*(2x)+4*1\\
6x-3x+3=8x+4\\
3x+3=8x+4 \quad | \quad -3x\\
3 = 8x-3x+4\\
3=5x+4 \quad | \quad -4\\
3-4=5x\\
-1=5x\\
5x=-1\quad | \quad :5\\
x=-\frac{1}{5}$$

For triangle HJK, j = 31, m∠H = 132 degrees, m∠J = 21 degrees, and m∠K = 27 degrees. Find h to the nearest whole number.

$$\small{\text{
$
\tan{(132)}=\frac{h}{p} \qquad
\tan{(27)}=\frac{h}{q} \qquad p+q=31\ m
$
}}\\
\small{\text{
$
p=\frac{h}{\tan{(132)}} \qquad
q=\frac{h}{\tan{(27)}} \qquad \frac{h}{\tan{(132)}} +\frac{h}{\tan{(27)}}=31\ m
$
}}\\
\small{\text{
$
h* \left(
\frac{1}{\tan{(132)}} + \frac{1}{\tan{(27)}}
\right)=31\ m
$
}}\\
\small{\text{
$
h* \left(
-0.90040404430 + 1.96261050551
\right)=31\ m
$
}}\\
\small{\text{
$
h*1.06220646121 =31\ m
$
}}\\
\small{\text{
$
h =\frac{31}{1.06220646121}\ m = 29.1845334520\ m
$
}}\\
\small{\text{
h to the nearest whole number:
$
h = 29\ m
$
}}$$

$$\sqrt{x+9}=9 \quad | \quad ()^2\\
x+9=81 \\
x=81-9\\
x=72$$

This one is the flop problem. you flop the last number when doing division. then it turns into multiplication

So for this it would be 9/8x5/4 so 9x5=45 and 8x4= 32 so your final answer is 45/32

If you wanted to make this into a mixed number, find out how many times 32 goes into 45. It goes in once so do 45-32=3 so basically your mixed number, simplified version is

1 3/32 and thats as far as you can get it.

Sorry, I know this answer was shown above, but the reason people go on this site is so they can get explanation, not cold hard calculator facts

find three consecutive integers so that three times the middle integer is five more than the sum of the first and third.

$$3a_2=5+a_1+a_3 \quad | \quad a_1 = a_2-1 \qquad a_3=a_2+1\\
3a_2=5+(a_2-1)+(a_2+1)\\
3a_2=5+a_2+a_2\\
3a_2=5+2a_2\\
a_2=5\\
a_1 = a_2-1 = 5-1=4\\
a_3=a_2+1 = 5+1=6$$

4, 5, 6

Woah, Sasini, that is really confusing, they ment simplifying not dividing

if you have 6/8, find a number that goes into both of them AKA 2

6/2 and 8/2 that is 3 and 4

so your answer is 3/4

This is how I always understood it,

It is just like +, First ignore the whole numbers, for now

now you have: 1/2-2/9

find a common denominator (The bottom number in a fraction)

how many times would you multiply 9 to make it even for 2 to go into? Twice, because that turns into 18. How many times do you need to multiply 2 to get 18? nine times. 

The rule is you have to do the same thing to both your Numerator (top number in a fraction) and your denominator, so if your multiplying 9x2 then you have to multiply 2x2 as well. If you do that you end up with 4/18 for the last number. Now if your multiplying the first number by 9. Then you have to multiply 1x9 and 2x9 which leaves you with 9/18

Now your equation is 9/18-4/18. Ignore the 18 for now, what is 9-4? 5 so your answer is 5/18, you cannot reduce that, so now on to your whole numbers:

9-1=8 so just add the two together and you get 8 5/18

I promise you, once you get the hang of it, its not as complicated as I made it seem

 I  expect you want me to factorize it?

$$\\7x^2+ 27xy + 18y^2\\$$

This is a tough one.  

We want to find 2 numbers that mult to 7*18  and add to 27

prime factors of 7*18  are  7*3*3*2   I notice straight off that 21+6=27

so the numbers are  21 and 6

change 27xy into    21xy+6xy

$$\\7x^2+ 27xy + 18y^2\\
=7x^2+ 21xy +6xy+ 18y^2\\
=7x(x+3y)+6y(x+3y)\\
=(7x+6y)(x+3y)$$

Do you understand Anima ?

3[9-(7-9)] Always do whats in the parenthesis 

3[9-(-2)] The two negatives together make a positive

3[9+2] Then do whats inside the parenthesis first

3[11] Then multiply the two together

3x11=33

33 is your answer (don't worry, I checked on a calculator)

Just subtract 7/16  from both sides.

$$\left({\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right){\mathtt{\,-\,}}\left({\frac{{\mathtt{7}}}{{\mathtt{16}}}}\right) = {\frac{{\mathtt{77}}}{{\mathtt{16}}}} = {\mathtt{4.812\: \!5}}$$

It is not infinity it is just too huge a number for the calc to display it.

Just subtract 2 from both sides

$$\\(y^{-5})^{10}y^{10}
=y^{-50}y^{10}
=y^{-40}
=\frac{1}{y^{40}}$$

that is a good answer frnd.

Yes

arctan is the same as inverse tan

so just use     atan               

6/8

 6 number / 2 and 8 number/2

3/4

that is the answer......

(2x - 3) ^ 2

4x^2 -12x +9

3 + 5 ^0

3 + 0

3//

21 laughing like h**l