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Please answer this!!!!!!!!!!!!!!!!!!!!!!!!!!!1

2^4 * 5^5 ==16 x  3125 ==50,000

50,000==[1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100, 125, 200, 250, 400, 500, 625, 1000, 1250, 2000, 2500, 3125, 5000, 6250, 10000, 12500, 25000, 50000]>>Total = 30 divisors.

EVEN perfect squares==4, 16, 100, 400, 2,500, 10,000==Total of 6

its urmomsucksonmeallnight

For the quadratic to be the square of a binomial, the discriminant(\(b^2 -4ac\)) must equal 0. 

Subsituting what we know, we have: \(28^2 - 4(16 + k) = 0 \). 

Now, we just have to solve for k. 

Can you take it from here?

We have the equation: \(\large{{{x +2} \over x} \times {x+3 \over x+1} \times {x+4 \over x+2} = 10}\)

Because there is the term \(x+2\) in both the numerator and the denominator, we can cancel them out.

This gives us: \(\large{{x+3 \over x+1} \times {x+4 \over x} = 10 }\)

Simplifying the left-hand side gives us: \(\large{{{x^2 + 7x + 12} \over {x^2 + x}} = 10 }\). 

From the equation, we know that the numerator (\(x^2 + 7x + 12 \)) must be 10 times the denominator (\(x^2 + x \))

Thus, we have: \(10(x^2 +x) = x^2 + 7x + 12 \)

Now, we have to solve the equation, and subsitute our values into the first fraction (\(\large{x+2 \over x}\))

Can you take it from here?

turn this into: 7x^2 + 8x - h = 0

using the quadratic formula to find the solutions:\({-8 \pm \sqrt{64+28h} \over 14}\)

this means that \(\sqrt{64+28h}\) = 0

h = \(-\)\(16\over 7\)

What is the ratio of the area of triangle AXY to the area of triangle ABC?

Hello Guest!

I assume: Right triangle, right angle in A. 

Then applies:

\(A_{AXY}:A_{ABC}=\frac{5\cdot 8}{2}:\frac{7\cdot10}{2}=\color{blue}4:7\)

If the angle in A is not 90°, I'm not sure if this ratio is the same.

  !

Prove triangle ABC is congruent to triangle DEF.

Hello Guest!

Triangles are congruent if they have the same three sides (sss).

\(BC = EF =+\sqrt{16+1}=+\sqrt{17}\\ AB=ED=2\\ AC=DF=+\sqrt{16+9}=5\)

  !

Oh I missed "even". Thanks. But the counting method is still the same.

a) The initial fixed charge is \(t_1 = 8\).

b) Charge per kilometre travelled is 0.75, because for each extra kilometer, the price is increased by 0.75, as described by the recursive formula.

c) This is an arithmetic sequence, since each term is 0.75 more than the previous one.

d) The general term of an arithmetic sequence is \(a + (n - 1)d\), where a is the first term and d is the common difference. Can you find those numbers and substitute into the expression? Don't forget to simplify the result.

e) After you have done d), substitute n = 16 into the expression also.

1      +      2   +  3+       4+.....2019+2020+2021+2022      flip it

2022+2021+2020+2019+......+    4+     3+      2+      1

add  it, which is then 2023 + 2023+ 2023+ 2023....., and there were 2022 of them in the first place, so multiply 2022 by 2023.

There are 2 sets of them, so divide by 2 because there was 1 in the question.

Note that when x = 4, y = 0. Then substituting (x, y) = (4, 0) gives:

\(\dfrac{4 + A}{4B + C} = 0\\ 4 + A = 0\\ A = -4\)

Now, note that the vertical asymptote occurs at x = 3. That means the denominator (Bx + C) is 0 when x = 3. Then 3B + C = 0.

Then, note that \(\displaystyle\lim_{x\to \infty} y = -1\) since y approaches -1 when x approaches infinity. Then \(\dfrac1B = -1\implies B = -1\).

Since 3B + C = 0, C = 3.

Therefore, the function is \(y = \dfrac{x - 4}{-x + 3}\). A + B + C = -4 + (-1) + 3 = -2.

First term = -56

Common difference = (-44) - (-56) = 12

Let n be the number of terms. Using the formula for arithmetic series,

\(\dfrac n2 (2(-56) + (n - 1)(12)) = 988\)

This is now a quadratic equation in n. Can you take it from here? Just note that n is a positive integer.

The pair is not unique. Here's the graph plotted in desmos:

https://www.desmos.com/calculator/uffp9pdojm

Let the total number of people on the ferry==T
0.75T ==adults
0.25T ==children

0.40 * 0.25T ==0.10T are girls
0.25T - 0.10T ==0.15T are boys.
0.75T - 0.15T ==168
0.60T ==168

T ==168 / 0.60 ==280 people on the ferry.
0.10 x 280 ==28 girls on the ferry
0.15 x 280 ==42 boys on the ferry

28 + 42 ==70 children on the ferry
280 - 70 ==210 adults on the ferry
70 / 0.40 ==175 people remained

280 - 175 ==105 adults left the ferry, or 1/2 of adults left the ferry.

210 - 105 ==105 adults remained on the ferry
105 adults + 70 children ==175 people remained on the ferry.

1+2+3+4+...+2019+2020+2021+2022==S, solve for S

S ==[2022 x 2023] / 2 ==2,045,253

There are 5 such palindromes as follows:

(2992, 3553, 4114, 7667, 8228) >>Total = 5

100 + 150 ==250 m/minute - their combined speed running towards each other.

400 / 250 ==1.6 minutes - when they will meet for the first time.

1.6 minutes x 3 ==4.8 minutes from the start time, when they meet for the 3rd time.

We have a right triangle and will use the Pythagorean Theorem to solve. 

The Pythagorean Theorem states: \(a^2 + b^2 = c^2 \), for any right triangle, where c is the hypotenuse, and a and b are legs. 

Substituting what we know, we have: \(20^2 = 12^2 + b^2 \)

Now, we have to solve for b and take the positive value. 

Can you take it from here?

6/6 x 5/6 x 4/6 x 3/6 x 2/6==6! / 6^5==5 / 54 - this is the probability that they are all DIFFERENT.

1 - 5/54 ==49 / 54 - probability of getting at least one matching pair.

(9, 8, 7, 6, 5, 4, 3, 2, 1) >>Total = 45

The above numbers are: number of ways of making up the numbers: 16, 17, 18, 19, 20, 21, 22, 23, 24

The probability is:  45 / 12^2 ==45 / 144==5 / 16

How many adults got off the ferry?

Hello Guest!

\(f=a+g+b\\ a=0.75(a+g+b)\\ 0.25f=b+g\\ a=b+168\\ 0.4(f-x)=g+b\\ 0.25f=0.4f-0.4x \)

\(0.15f=0.4x\\x=0.375f\\ f=b+168+b+g\\ 4b+4g=2b+g+168\\2b+3g=168\)

to be continued

\(\)

Find the value of
\(S_n=\dfrac{1}{2} + \dfrac{3}{2^2} + \dfrac{5}{2^3} + ... + \dfrac{2n-3}{2^{n-1}} + \dfrac{2n-1}{2^n}\)

\(\text{Let $x=\dfrac12$}\)

\(\begin{array}{|rcll|} \hline S_n&=&x + &3x^2 + 5x^3+ 7x^4 + ... + (2n-3)x^{n-1} + (2n-1)x^n \\ xS_n&=& &1x^2 + 3x^3+ 5x^4 + ... + (2n-5)x^{n-1} + (2n-3)x^n+(2n-1)x^{n+1} \\ \hline x-xS_n &=& x + &2x^2 + 2x^3+ 2x^4 + ... + 2x^{n-1} + 2x^n -(2n-1)x^{n+1} \\ S_n(1-x) &=& x + &2*(x^2 + x^3+ x^4 + ... + x^{n-1} + x^n) -(2n-1)x^{n+1} \\ &&& \boxed{x^2 + x^3+ x^4 + ... + x^{n-1} + x^n = \dfrac{x^2-x^{n+1} } {1-x} } \\ S_n(1-x) &=& x + &2*\dfrac{x^2-x^{n+1} } {1-x} -(2n-1)x^{n+1} \quad | \quad 1-x=1-\dfrac12=\dfrac12 \\ S_n*\dfrac12 &=& x + &4(x^2-x^{n+1}) -(2n-1)x^{n+1} \quad | \quad * 2 \\ S_n &=& 2x + &8(x^2-x^{n+1}) -2(2n-1)x^{n+1} \\ S_n &=& 2x + &8x^2-8x^{n+1} -2(2n-1)x^{n+1} \\ S_n &=& 2x + &8x^2-2x^{n+1}\Big(4 +(2n-1)\Big) \\ S_n &=& 2x + &8x^2-2x^{n+1}(2n+3) \\ S_n &=& 2x \Big(& 1 + 4x-x^n(2n+3)\Big) \quad | \quad x=\dfrac12 \\ S_n &=& &1 + 2-\dfrac{2n+3}{2^n} \\ \mathbf{S_n} &=& \mathbf{3-}&\mathbf{\dfrac{2n+3}{2^n}} \\ \hline \end{array} \)