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Let a1 be the first term

Let d be the difference between each term (that is, a2 - a1, or a3 - a2, etc)

a1 + (a1 + 2d) = -2

(a1 + d) + (a1 + 3d) + (a1 + 5d) = 1

Simplify first equation:

2a1 + 2d = -2

a1 + d = -2

Wait what? we see a1 + d = a2 = -2 !!!

Simplify second equation

-2 + (2a1 + 8d) = 1

2a1 + 8d = 3

minus 2(a1 + d) = 2(-2)

6d = 3 - (-4)

d = 6/7

since a2 = -2, 

d = -2 - 6/7 = -2 6/7 = -20/7

The sum of the angles of the quadrilateral = 360 degrees ....

so 

 q   +    4q+17     +   8q+21     +    15q+148 = 360 

28q + 186 = 360 

q = 6.214               Then the largest angle is  15 (6.214) + 148 = 241.2 degrees 

First , simplify the equations :

s/2   + 5t = 3 - 7t + 8s 

   =    S + 10T = 6 - 14T + 16S

     =  24T = 15S + 6 

3T - 6S = 9 - 2T

= 5T = 6S + 9         Multiply this equation by -2.5 to get 

  -12.5 t = -15s - 22.5      And add this to the first equation

   11.5 t =  - 16.5

        T = - 16.5/11.5  = - 33/23       Use this value in one of the equations to solve for s :      S =  - 62/23 

Call "R" the addition between terms  

From T₂ to T₅ is 3R and T₅ is 5 less than T₂    

From T₅ to T₈ is 3R  so  T₈ is 5 less than T₅   

T₈ is 12 – 5 so T₈ = 7    

_.   

The answer is

\(s=-\frac{62}{23}\\ t=-\frac{33}{23}\)

First, let's use the Law of Cosines. It states that \(cos PMQ = - cos PMR\)

Thus, we can write the equations

\(PQ^2 = QM^2 + PM^2 - 2(QM * PM) * (-cos PMR) \\ PR^2 = RM^2 + PM^2 - 2 (RM * PM) * ( cos PMR)\)

Plugging in the values we already know from the problem, we get

\(5^2 = 5.5^2 + PM^2 + 2(5.5 * PM)cos(PMR) \\ 8^2 = 5.5^2 + PM^2 - 2(5,5 * PM) cos (PMR)\)

Now, add these two equations. We get

\(5^2 + 8^2 = 2 * 5.5^2 + 2PM^2 \\ 89 = 60.5 + 2PM^2 \\ 28.5 / 2 = PM^2 \\ 14.25 = PM^2\)

Thus, we have \(\sqrt{14.25} = PM ≈ 3.77\)

The answer is 3.77. 

Thanks! :)

We can use the great pythaogrean theorem to our advantage. 

First off, let's set some variables. 

Let's let BC = a and AC = b. 

Now, using pythagoras, let's right some equations. We have

\(\overline{AM} = \sqrt{b^2 + \left(\dfrac a2\right)^2} = \dfrac12 \sqrt{4b^2 + a^2}\) since AM is the median and we also have

\(\overline{BN} = \sqrt{a^2+ \left(\dfrac{b}{2}\right)^2} = \dfrac12 \sqrt{4a^2 + b^2}\) since BN is also the median of the triangle. 

Now, we're trying to isolate a^2+b^2 because that's how we find the hypotenuse of the triangle. 

Thus, we now have

\(4b^2 + a^2 = 2^2 \\4a^2 + b^2 = 2^2\)

Now, adding the two equations up with each other and dividing both sides by 5, we find that

\(a^2 + b^2 = \dfrac{4+4}{5} = 8/5\)

So the hypotenuse is that square rooted, which is \(\sqrt{\frac{8}{5}}\)

So \(\sqrt{\frac{8}{5}}\) is our final answer. 

Thanks! :)

See the image attached from EP. 

The answer is 9.25. 

Thanks! :)

We can set variables up to solve this problem. Let's set 

From the problem, we can write the equations

\(S_5 = 5a_1 + (d + 2d + 3d + 4d) = 5a_1 + 10d\\S_{10} = 10a_1 + (d + 2d + \cdots + 9d) = 10a_1 + 45d\\\\\S_{15} = 15a_1 + (d + 2d + \cdots + 14d) = 15a_1 + 105d\)

Simplifying the first 2 equations, we get the conditions

\(\begin{cases} 5a_1 + 10d = \dfrac15\\ 10a_1 + 45d = \dfrac1{10} \end{cases}\)

Solving this gives that d is -3/250 and a_1 is 8/125. 

Thus, we have

\(S_{15} = 15a_1 + 105d = -\dfrac3{10}\)

So -3/10 is our answer. 

Thanks! :)

the sequence goes as following:

2, 7, 12, 17, ...

the sequence of sums goes:

2, 9, 11, 28, 50, 77, 105

The seventh term sends Shirley running from the room.

That number is 2+6*5=32

For a system of equations to have infinitely many solutions, the equations should be the same. We start by simplifying:

\(8a+b=-3\)

\(9a+(2-m)b=k\)

now we make the a terms the same:

\(8a+\frac{8}{9}*(2-m)b=\frac{8k}{9}\)

this means

 \(\frac{8k}{9}=-3, so\: k=-\frac{27}{8}\)

and

\(2-m=1, so\:m=1\)

Understanding the Problem

We're given an inverse relationship between mice and cats:

Mice = k/cats

We have two sets of data:

When cats = 3r - 19, mice = 2r + 1

When mice = 6r - 24, cats = r - 3

We need to find the value of k.

Setting up the Equations

Using the given information, we can create two equations:

2r + 1 = k / (3r - 19)

6r - 24 = k / (r - 3)

Solving for k

We can solve either equation for k and then substitute it into the other equation. Let's solve the first equation for k:

k = (2r + 1)(3r - 19)

Now, substitute this expression for k in the second equation:

6r - 24 = [(2r + 1)(3r - 19)] / (r - 3)

Simplify the equation:

(6r - 24)(r - 3) = (2r + 1)(3r - 19)

6r^2 - 42r + 72 = 6r^2 - 35r - 19

7r = 91

r = 13

Now, substitute r = 13 back into either of the original equations to find k. Let's use the first equation:

k = (213 + 1)(313 - 19)

k = (27)(20)

k = 540

Therefore, the value of k is 540.

Applying the binomial formula and distributing everything out (I have no life), we get

\(27x^{3}+81x^{2}y+81xy^{2}+27y^{3}+81x^{2}+162xy+81y^{2}+81x+81y+27 \)

As shown, 162 is the coefficient. 

Thanks! :)