All Answers 356464 Answers

thank you so much!

Yes. 225 is the correct result.

Yes that's exactly what I meant, I tried figuring it out with the help of YouTube videos and I ended up with   https://snag.gy/qGM8Vn.jpg

Is that correct?

If you mean  \(\sum_{n=1}^{15}(2n-1)\) then the following should help:

Can you take it from here?

(1004, 1015, 1026, 1037, 1048, 1059, 1070, 1081, 1092, 1103, 1114, 1125, 1136, 1147, 1158, 1169, 1180, 1191, 1202, 1213, 1224, 1235, 1246, 1257, 1268, 1279, 1290, 1301, 1312, 1323, 1334, 1345, 1356, 1367, 1378, 1389, 1400, 1411, 1422, 1433, 1444, 1455, 1466, 1477, 1488, 1499, 1510, 1521, 1532, 1543, 1554, 1565, 1576, 1587, 1598, 1609, 1620, 1631, 1642, 1653, 1664, 1675, 1686, 1697, 1708, 1719, 1730, 1741, 1752, 1763, 1774, 1785, 1796, 1807, 1818, 1829, 1840, 1851, 1862, 1873, 1884, 1895, 1906, 1917, 1928, 1939, 1950, 1961, 1972, 1983, 1994) = 91 - Numbers

Note: It starts at 1004 and then every 11 numbers after that give a remainder of 3

so its -4/4 = -1 and so on?? 

So for question 1, which choice would it be? And for question 2, would it be the 3rd choice or which choice would it be?

Also, is my choice for the below question correct?

You know you can just copy and pate it and through it in the froum calaculator it tracks symbols.

is it 23 thats what it said in the calaculator.??? I'm not interarly sure you intended this to be madiatory or your just practicing pemdas.

Try using x = 0,1,2,3 etc   and see what f(x)  and g(x) are 

see when they give the same answer :)

Thanks Omi67!

Problem 1

The formula for the volume of a cone is 𝑉=< =𝜋𝑟(ℎ.  What is 𝑟 in terms of 𝑉 and ℎ? 

Problem 2

Solve the following system of equations: 

2𝑥+2𝑦=250 𝑥𝑦=600

Problem 3

Graph and solve the following system of inequalities: 

20𝑥+50𝑦<250 𝑦≤−0.25𝑥(+2.5𝑥)

Inverse Functions

If
\(\large{\mathbf{f(x)=\dfrac{a}{x+2}}}\),

solve for the value of  \(a\) so that

\( \large{\mathbf{f(0)=f^{-1}(3a)}}\).

\(\begin{array}{|rcll|} \hline f\Big(f^{-1}(x) \Big) &=& x \quad | \quad x = 3a \\ f\Big(f^{-1}(3a)\Big) &=& 3a \quad | \quad f^{-1}(3a) = f(0) \\ f\Big(f(0)\Big) &=& 3a \quad | \quad f(0) = \dfrac{a}{0+2} = \dfrac{a}{2} \\ f\Big(\dfrac{a}{2}\Big) &=& 3a \quad | \quad f\Big(\dfrac{a}{2}\Big) = \dfrac{a}{\dfrac{a}{2}+2} = \dfrac{2a}{4+a} \\ \dfrac{2a}{4+a} &=& 3a \quad | \quad : a \\ \dfrac{2}{4+a} &=& 3\\ 3(4+a) &=& 2 \\ 12+3a &= 2 \\ 3a &=& -10 \\ \mathbf{a} &\mathbf{=}& \mathbf{-\dfrac{10}{3}} \\ \hline \end{array} \)

Find constants A, B, and C so that

\(\large{\dfrac{4x}{(x - 5)(x - 3)^2} = \dfrac{A}{x - 5} + \dfrac{B}{x - 3} + \dfrac{C}{(x - 3)^2}}\)

\(\begin{array}{|lrcll|} \hline & \dfrac{4x}{(x - 5)(x - 3)^2} &=& \dfrac{A}{x - 5} + \dfrac{B}{x - 3} + \dfrac{C}{(x - 3)^2} \quad | \quad \cdot (x - 5)(x - 3)^2 \\\\ & \mathbf{4x} &\mathbf{=}& \mathbf{A(x - 3)^2 + B(x - 5)(x - 3) + C(x - 5)} \\\\ \hline x=3: & 4\cdot 3 &=& A\cdot 0 + B\cdot 0 + C(3 - 5) \\ & 12 &=& -2C \\ & \mathbf{C} &\mathbf{=}& \mathbf{-6} \\\\ x=5: & 4\cdot 5 &=& A(5 - 3)^2 + B\cdot 0 + C\cdot 0 \\ & 20 &=& A\cdot 2^2 \\ & 20 &=& 4A \\ & \mathbf{A} &\mathbf{=}& \mathbf{5} \\\\ x=0: & 4\cdot 0 &=& 5(0 - 3)^2+B(0 - 5)(0 - 3)-6(0 - 5) \\ & 0 &=& 45 +15B+30 \\ & 15B &=& -75 \\ & \mathbf{B} &\mathbf{=}& \mathbf{-5} \\ \hline \end{array}\)

I think you meant to put (2+p)i + (5+q)j   (you have q in both terms!).

Anyway, here is a solution

Binomial Expansion  approach

(6701)^83  =  

(6700 + 1)^83  =  6700^83  + C(83,1)*6700^82 +  ...+ C(83,81)*6700^2 + C(83,82)*6700 + 1

Note that every term in red will have at least 4 trailing zeros, so  each of these terms is divisible by 1000

So....we only need consider the last two terms

C(83, 82) * 6700  +  1     =

83 * (6700)  + 1  =

(80 + 3) * (6700)  + 1 =

80*6700 + 3*6700  + 1 

The first term  will have 3 trailing zeros, so....it is divisible  by 1000....and

3 * 6700 + 1 = 20100  + 1 =  20(1000) +  100  +  1

So  we have  left

 [ 100 + 1 ] / 1000   ⇒    remainder 101