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Solve the following system:
{3 + 10 d + 5 n = 168 | (equation 1)
5 + 10 d + 5 (n - 1) + p = 168 | (equation 2)
d + n + p = 24 | (equation 3)

Express the system in standard form:
{5 n + 10 d+0 p = 165 | (equation 1)
5 n + 10 d + p = 168 | (equation 2)
n + d + p = 24 | (equation 3)

Subtract equation 1 from equation 2:
{5 n + 10 d+0 p = 165 | (equation 1)
0 n+0 d+p = 3 | (equation 2)
n + d + p = 24 | (equation 3)

Divide equation 1 by 5:
{n + 2 d+0 p = 33 | (equation 1)
0 n+0 d+p = 3 | (equation 2)
n + d + p = 24 | (equation 3)

Subtract equation 1 from equation 3:
{n + 2 d+0 p = 33 | (equation 1)
0 n+0 d+p = 3 | (equation 2)
0 n - d + p = -9 | (equation 3)

Swap equation 2 with equation 3:
{n + 2 d+0 p = 33 | (equation 1)
0 n - d + p = -9 | (equation 2)
0 n+0 d+p = 3 | (equation 3)

Subtract equation 3 from equation 2:
{n + 2 d+0 p = 33 | (equation 1)
0 n - d+0 p = -12 | (equation 2)
0 n+0 d+p = 3 | (equation 3)

Multiply equation 2 by -1:
{n + 2 d+0 p = 33 | (equation 1)
0 n+d+0 p = 12 | (equation 2)
0 n+0 d+p = 3 | (equation 3)

Subtract 2 × (equation 2) from equation 1:
{n+0 d+0 p = 9 | (equation 1)
0 n+d+0 p = 12 | (equation 2)
0 n+0 d+p = 3 | (equation 3)

n = 9 Nickels
d = 12 Dimes
p = 3 Pennies

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Hi Shreyas1,

You most likely would have had an answer much more quickly if your question was presented in a readable fashion!

But here is your answer  

\(\frac{BD}{DC}=\frac{4}{3}\\ \text{so let BD=4x and DC=3x }\\ Area\;\;\triangle ABD=0.5*4x*h=24\\ \qquad \qquad 2xh=24\\ \qquad \qquad xh=12\\~\\ Area\;\;\triangle ABC=0.5*7x*h\\ Area\;\;\triangle ABC=3.5xh\\ Area\;\;\triangle ABC=3.5*12\\ Area\;\;\triangle ABC=42cm^2\\\)

NOW do the subtraction to get the final answer :)

Kinda

pls can someone help

I have been waiting for someone to help me for a long time

Like this:  \(\frac{(4x^3y^2)^3}{10}\rightarrow\frac{4^3(x^3)^3(y^2)^3}{10}\rightarrow\frac{64x^9y^6}{10}\rightarrow\frac{32x^9y^6}{5}\)

This doesn't involve proving trig identities, but using them to find the value(s) of \(\theta\)  and x that make the equations true.

By definition: \(\sec\theta=\frac{1}{\cos\theta}\)

Using this in a and rearranging:   \(\frac{1}{\cos^2\theta}=2\tan\theta\)

Multiply both sides by \(\cos^2\theta\) and using the fact that \(\tan\theta=\frac{\sin\theta}{\cos\theta}\) we get, on simplifying: \(1=2\sin\theta\cos\theta\)

The RHS is just \(\sin{2\theta}\)  so we have \(1=\sin{2\theta}\)

This results in \(2\theta=\frac{\pi}{2}\text{ or }\theta=\frac{\pi}{4}\) 

For b rewrite x as \(2\theta\) and use \(\cos{2\theta}=\cos^2\theta-\sin^2\theta\text{ and }\sin{}2\theta=2\sin\theta\cos\theta\text{ together with }\cos^2\theta+\sin^2\theta=1\)

You should eventually find there are two values of x between 0 and 2pi that satisfy the equation.

Remember that \(\cos^2x+\sin^2x=1\)

Rearrange this in the form:  \(\sin^2x=1-\cos^2x\)

Substitute this into the left hand side (LHS) of your identity:

LHS:  \(1-\cos^2x-cos^2x\)

Simplify

LHS: \(1-2\cos^2x\)

So LHS = RHS

.

Thanks gamesmaster

\(\sqrt{x^2}=|x| \qquad not\;\;x\)

If this expression is real then y must be positive.   

Also if an expression is simple then the denominator should be rational. (no Square root on the bottom)

Although your teacher may not care about this.\(\)

\(\sqrt\frac{4x^2}{3y}\\=\frac{|2x|}{\sqrt{3y}}\\ =\frac{|2x|}{\sqrt{3y}}\times \frac{\sqrt{3y}}{\sqrt{3y}}\\ =\frac{|2x|\sqrt{3y}}{3y} \\ =\left| \frac{2x\sqrt{3y}}{3y} \right|\\~\\ = \frac{2x\sqrt{3y}}{3y} \text{ when }x\ge0\\ = \frac{-2x\sqrt{3y}}{3y} \text{ when }x<0\)

Presumably solutions means solutions to f(x) = 0

In this case there are 3 unique solutions, you can read them right off as

x = 3, x = -2, x = 1

in (2)

a, b, and c are the values for which f(x) = 0, i.e. crosses the x axis

f(x) = (x+2)(x-2)(x-6)

f(x)= (x+1.464)(x-2)(x-2)

The roots are where the function = 0

f(x) = (x-4)(x-7)

I'll start you out....you can finish it, I'm sure....

City 1 :

Month 1 =12

Month 2 = 18

           3 = 21  (I think)

           4 = 16

           5 = 10

Month 6 =  6         Add these numbers and divide by 6 to get the mean

     12 + 18 + 21 + 16 + 10 + 6    = 83      83/6 = 13  5/6   inches = mean

Now do this for City 2    and compare......

3  3/8 = 27 /8

27/8  /  3/8   =  27/8 X 8/3 = 9  poles

6 2/3  /   1 1/2   =   20/3  /   3/2    =    20/3  X  2/3   = 40/9 =  4  4/9 containers

All the points on the green lines satisfy the first condition

 and

All the points on the orange lines satisfy they second condition.

All the points that are on both the green and the orange lines satisfy both conditions.

I have labelled 4 of the poiints but I missed one.  Which one did I miss ?    

What you have here is a locus (set) of points such that  

x=2t+3      and      y=3-3t   for t is any real value.

so what you need to do it make t the subject of both these equations

So now you know that 

\(t=\frac{x-3}{2}=\frac{3-y}{3}\\ \frac{x-3}{2}=\frac{3-y}{3}\\ 6 \cdot \frac{x-3}{2}=6 \cdot \frac{3-y}{3}\\ 3(x-3)=2(3-y)\\ 3x-9=6-2y\\ 3x+2y-15=0 \qquad \text{which is a line}\\ or\\ 2y=-3x+15\\ y=\frac{-3x}{2}+\frac{15}{2}\)

which is still the same line.      gradient = -1.5    and    y intercept = -7.5

y=(x-(-3) ) (x-2)

y = f(x) = x^2 +x -6

The x-intercept occurs when y= 0   so put y= 0 in to the equation

5x-12(0) = 60

x = 12

Similarly the y-intercept occurs when x =0

5(0) -12y = 60

y=5

Distance between =   sq rt (12^2 + 5^2) = 13       (Pythagorean theorem)

I think this goes like this:

11/13 - 9/11 = The common denominator of 11 and 13 is 143

[ 121 - 117] / 143 =

4 / 143

You are multiple of 8 such as the numbers you have listed. 104, 112, 120, 128, 136, 144. They form an arithmetic series whose first term is 104, and the common difference is 8. In other words: 104+8=112. 112 + 8=120. 120 + 8=128. 128 + 8 =136. 136 + 8 =144. And that is it.

Let the integers be:

n,  n + 2,  n  + 4

2n + (n + 2) - 1 + (n + 4) + 4 = -15

4n - 1 + 10 = -15

4n + 9 = -15

4n = -15 - 9

4n = - 24

n = - 6 - the smallest number

-6 + 2 = - 4 - the middle number.

-6 + 4 = - 2 - the largest number.

Are you sure you have your expressions written out correctly?? As they are, they both give 9 as the answer and 9/9 =1.