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Expanding the quadratic, we get: \(2x^2+16x+30 <3x^2+63x+162\)

Bringing everything to the left-hand side, we get: \(-x^2-47x-132<0\)

Multiply through by -1 to make all the coefficients positive: \(x^2+47x+132>0\) (Remember to switch the sign, because we multiplied by a negative!!)

Factoring this equation, we get: \((x+44)(x+3)>0\)

For the result to be positive, we either need: \(\text{pos} \times \text{pos}\) or \(\text{neg} \times \text{neg}\).

You can do the rest from here, right?

The area of the circle is \(\pi r^2 = 100 \pi\).

We know that the win sector is \(2 \over 5\) of the entire circle, thus the area of the win sector is: \({2 \over 5 }\times 100 \pi = \text{_____}\)

The smallest multiple of 7 end in 8 ==28

The largest multiple of 7 ending in 8=1988

Every number beginning with 28 + 70 ==multiple of 7 ending in 8

[1988 -  28] / 70  + 1 ==29 multiples of 7 that end end in 8 between 1 and 2000.

There are infinitely many pairs of solutions for (x, y).

I know the answer, but I don't know how to do it quickly. 

If two quadratic polynomials have the exact same roots, then one of them is a multiple of the other.

Judging by the coefficients of x, Ax^2 + x + B should be 1/8 times of x^2 + 8x + 4.

You can continue by expanding \(\dfrac18 (x^2 + 8x + 4)\) and finding the coefficient of x^2 and the constant term.

Since it is a translation, \(AA'B'B\) is a parallelogram or a straight line.

The diagram is shown below:

From the diagram, you can see that BA' = 12.

Let the number of 1-dollar coins==D

Then the number of 20-cent coins==5/7D

Let the numbe of withdrwals ==N

D - 6N == 4...................(1)

5/7D - 4N ==7...............(2), solve for D, N

D ==91 - number of 1-dollar coins that he started with

5/7 x 91 ==65 - number of 20-cent coins that he started with.

91 - 65 ==26 - more 1-dollar than 20-cent coins in his piggy bank at the start.

Hint: If a number \(N\) in base 10 has 4 digits when written in base 3, that means \(27 \leq N < 81\), because 27 is the least number to have 4 digits in base 3 (which is 1000), and 81 is the least number to have 5 digits in base 3 (which is 10000).

Also, if it has 3 digits when written in base 5, \(25 \leq N < 125\) for a similar reason.

If an integer satisfies both inequalities, that means \(27 \leq N < 81\).

I believe you can continue on your own.

But in case you are still clueless, feel free to tell me.

Let \( d = \deg (p) \in \mathbb Z^+\).

Note that \(\deg(p\circ p) = d^2\) and \(\deg(x\cdot p + x^2 + 3x + 1) = \max(d + 1, 2)\)

Now we eliminate the case when p(x) is constant to ensure that \(\max(d + 1, 2) = d + 1\).

(Because if p(x) = c, c = cx + x^2 + 3x + 1 for any x, which is impossible.)

Comparing the degrees on both sides gives \(d^2 = d + 1\), which has no solution in \(\mathbb Z^+\).

Therefore, there are no such polynomial p(x) with p(p(x)) = xp(x) + x^2 + 3x + 1.

The answer is \(y=2x-3.\)

\(45\) degrees

Explanation: https://web2.0calc.com/questions/help-please_11085

\(116\)

The first card can be any suit so we don't need to include it into the calculation.

Because each card is replaced after it is drawn, there is a 1/4^th chance that the second card will match the first.

It is also true that the third card has a 1/4^th chance that it will match the first two.

Lastly, the fourth card has a 1/4^th chance that it will match the first three.

Multiplying together these possibilities, we get a probability of (1/4) x (1/4) x (1/4)  =  1/64.

He placed the following amount on square #18

2^(18 - 1) ==2^17 ==131,072 - grains of wheat

[2^64] - 1 ==18,446,744,073,709,551,615 - total number of grains of wheat on the chess board.

18,446,744,073,709,551,615 x 1 / 7000 ==2,635,249,153,387,078.80 - pounds of grain, or their total weight.

Let the total distance of the journey==D

1  - 3/5D ==2/5D -this equals the remaining 84 km

2/5D == 84    Cross multiply

2D ==84 x 5 ==420, solve for D

D ==420 / 2==210 km - total distance of his journey.

Average speed ==Total distance / total time

                           == 210 / 3 ==70 km/hour -his average speed for the entire trip.

We have the equation: \({{9+x+2x-3+7} \over 4}= 7\)

Now, solve for x, then compare the values...

Write as: \({p^3 \times q^9 \times 4^2 \times p^4 \times q^4} \over 8p^3 \times q^9\)

Combining like terms, we get: \({16 \times p^7 \times q^{13}} \over {8 \times p^3 \times q^9}\)

To divide the remaining exponents, you take the power of the greater number and subtract that from the smaller number. For example, \(q^{13} \div q^9 = q^{13-9} = q^4\). Now, you have to do these to the remaining terms.