6/17 ==0.3529411764705882 3529411764705882
The fraction ha a "period" of 16 digits, which means the first 16 digits over and over again.
Therefore: 100 mod 16 ==4 - or the fourth digit from the decimal, which is a "9".
30000 * 4/5 * 0.9 *0.9 *0.9
The sum of the series is 32.
Hey, if you try and list all the possible additions --- you might miss some...
We want to try to minimize the cases to minimize mistakes. We have 4 cases:
Case1 = no touchdowns
Case2 = 1 touchdown
Case3 = 2 touchdowns
Case4 = 3 touchdowns
Case4: If there are 3 touchdowns, then that means there is one possible way since you must have 3 touchdowns and one field goal.
Case4 = 1 way
Case3: If there are 2 touchdowns, we are left with 10 points. 3x + 2y = 10. If x = 0, then y = 5. If x = 1, then y is invalid. If x = 2, then y = 2. If x = 3, y is invalid. Thus, there are 2 ways if for Case3.
Case3 = 2 ways
Case2: If there is 1 touchdown, we are left with 17 points. 3x + 2y = 17. If x = 0, then y is invalid. If x = 1, then y = 7. If x = 2, then y is invalid. If x = 3, then y = 4. (We notice that x has to be odd) If x = 5, then y = 1. If x = 7, then y is negative --- impossible...
Case2 = 3 ways.
Case1: If there are no touchdowns, we have to make do with 24 points, 3x + 2y = 24. x = 0, y = 12. (x must be even) x = 2, y = 9. x = 4, y = 6. x = 6, y = 3. x = 8, y = 0. Case1 has 5 cases (x = 0, 2, 4, 6, 8).
Case1 = 5 ways.
Hence, we have 1 + 2 + 3 + 5 ways = 11 ways.
proyaop :)
+1
3T =4[T + 8/60], solve for T
The above line should read:
3T =4[T - 8/60], solve for T
Let her time in hours==T
T ==8/15 x 60 ==32 minutes - she was walking when you caught up.
32 - 8==24 minutes that you walked.
It is literally just (x+3)(x+5).
x²+8x+15.
Thanks, but thats wrong
Thank you. Just for clarification, is there a reason why it's 2 and not the others? Does it have something to do with the decimals?
The point (1,6) must be on the graph of y = f(x + 4).
There is only one solution, x = 8/3.
\(g_{43} = 108\)
We let \(b = a^2\).
We can simplfy the equation to: \(b^2 -3b+2\).
To factor, we need a pair of numbers that multiply to 2 and add to -3.
We see that the only pair satisfying this is -2 and -1.
Now, we can rewrite as: \(b^2 - b + 2b + 2\).
Factoring the first 2 terms and the last 2 terms seperately gives us: \(b(b-1) - 2(b-1)\).
We can add these to get: \((b-2)(b-1)\).
Can you take it from here?
n==33 , 66 , 99 , 132 , 165 , 198 , 231 , 264 , 297 , 330 , 363 , 396 , 429 , 462 , 495 , 528 , 561 , 594 , 627 , 660 , 693 , 726 , 759 , 792 , 825 , 858 , 891 , 924 , 957 , 990 , Total = 30 such values of n
Simplify this a little:
f(x) = 3x^2 - 4x - x^2 + 7x + 8 = 2x^2 + 3x + 8
This is a quadratic function, which means its graph is a parabola. The axis of symmetry of the graph is \(x = - \dfrac{3}{2(2)} = -\dfrac34\). (It is always x = -b/(2a) for the parabola y = ax^2 + bx + c.)
That means \(f(x) = f\left(2\left(-\dfrac34\right) - x\right) = f\left(-\dfrac32 - x\right)\). The required constant is -3/2.
Converting back to multiplication gives:
\(\quad \dfrac{\sqrt{49x}}{\sqrt{21y}} \div \dfrac{\sqrt{8z}}{\sqrt{xyz}} \div \dfrac{\sqrt{338x}}{\sqrt{18y}}\\ =\dfrac{\sqrt{49x}}{\sqrt{21y}} \times \dfrac{\sqrt{xyz}}{\sqrt{8z}} \times \dfrac{\sqrt{18y}}{\sqrt{338x}}\\ = \dfrac{21\sqrt 2 xy \sqrt z}{52 \sqrt{21}\sqrt{xyz}}\\ = \dfrac{\sqrt{42xy}}{52}\)
(a) By Cauchy-Schwarz inequality,
\((c^4 + d^4)(e^4 + f^4) \geq ((ce)^2 + (df)^2)^2\)
Then, again, by Cauchy-Schwarz inequality,
\(\begin{array}{rcl} (a^2 + b^2)^2(c^4 + d^4)(e^4 + f^4) &\geq & (a^2 + b^2)^2((ce)^2 + (df)^2)^2\\ &=& \left((a^2 + b^2)\left((ce)^2+ (df)^2\right)\right)^2\\ &\geq& ((ace + bdf)^2)^2\\ &=& (ace + bdf)^4 \end{array}\)
For (b), you can instead prove that \((a^2 + b^2)^2 (c^2 + d^2)^2 (e^2 + f^2)^2 \geq (ace + bdf)^4\). The rest follows from the fact that \((x + y)^2 \geq x^2 + y^2\) for nonnegative x, y, and part (a). If you can prove this inequality, the result of part (b) immediately follows.
Thanks!
Would this be right for d?
8 + (n - 1)0.75
= 8 + 0.75n - 0.75
= 0.75n + 7.25
Is there a reason for e) I would substitute n = 16 for a 15km trip and not n = 15?
The method hipie gave is also possible, but it is tedious. Instead, we could simplify the square roots on their own first.
\(\quad \dfrac{\sqrt{160}}{\sqrt{252}} \times \dfrac{\sqrt{245}}{\sqrt 3}\\ =\dfrac{4 \sqrt{10}}{6 \sqrt 7} \times \dfrac{7 \sqrt 5}{\sqrt 3}\\ = \dfrac{4\times 7}{6} \times \sqrt{\dfrac{10 \times 5}{7 \times 3}}\\ = \dfrac{4\times 7 \times \sqrt{10 \times 5 \times 7 \times 3}}{6\times 7 \times 3}\\ = \dfrac{2\sqrt{1050}}{9}\\ = \dfrac{10 \sqrt{42}}{9}\)
I have made a GeoGebra simulation of the problem, and it appears that the curve that H traces is an ellipse with major axis 4 units and minor axis 2 units.
GeoGebra simulation: https://www.geogebra.org/calculator/my8uamwy
Therefore, the area inside the curve is \(\pi \cdot 4 \cdot 2 = 8\pi\text{ square units}\).
Yes, you're right. It is because (1 + x)^10 (1 - x)^10 is just (1 - x^2)^10.
An integer k has inverse modulo 9 if gcd(k, 9) = 1.
Checking one by one, the integers that have inverse modulo 9 are 1, 2, 4, 5, 7, 8.
\(\sqrt{9-5+5-9+5-9+5} = \sqrt{1} = 1\) by evaluating the expression inside the square root.
