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Using similar triangles, TRQ is similar to TPS with an area ratio of 1/16 so the area of PQRS is 405/17

:D

Using Simons favorite factoring trick gets us (p-2)(q-2)=6 so (p-2) can be either 1,2,3, or 6 meaning p is 3,4,5, or 8. 2/5

:D

2021 is 6 mod 21 and the powers of 6 repeat in the infinite loop 6-5-6-5-6-5... and 2021 is 1 mod 2 so 2021^2021 is 6 mod 31.

:D

I can't understand your latex but try finding a value of a so that the range of f(x) is \((-∞,∞)\)

:D

Maybe this will help...

So first you can tell that the pattern is times \(1\over6\) because 9 divided by 3/2 = 6.

So you get \(9\)\(\times \) \(1\over6\)⁹=6^9 =\(9\) \(\times\)\(1\over10077696 \)=\(1\over1119744\).

So the answer is \(1\over1119744\).

-42

The negative of an absolute value is always negative

EEEEEEEE

There are 2 divisors(1,2)

Because subtracting an even number from another even number will always result in an even number, and 2a always has to be even. The only way fo 3b to be even is if b is even too. So b is any even number. 2b+30 can only guarantee the positive divisor 2 because we can find examples where it will not work for the other integers. Since b is and integer, 1 is already a lock. 

b=4 a=-2

3 wont work, 4 wont work, 5 wont work and 6 wont work.

Good job CoolStuffYT!

-Grace

Let

 \(\displaystyle p(x)=x^{4}+ax^{3}+bx^{2}+cx+d, \\ \text{and} \\ q(x)=p(x)-x^{4}.\)

q(x) is a cubic, so its third difference column will be constant.

\(\displaystyle q(1)=p(1)-1^{4}=1-1=0, \\q(2)=p(2)-2^{4}=2-16=-14, \\ q(3)=p(3)-3^{4}=3-81=-78, \\ q(4)=p(4)-4^{4}=4-256=-252.\)

The third difference turns out to be -60, and running that down the diagonal leads to q(5) = -596.

From that, 

\(\displaystyle -596=p(5)-5^{4}, \\ p(5)=625-596=29.\)

Assuming Shamique is asking the question, then, in equation form the information is:

S = 2b           (1)              where S is Shamique's age, b is her brother's age

S = s - 5        (2)              where s is her sister's age

s + 5 = 2(b + 5)              However, this is not independent information as it can be derived from equations (1) and (2), so we ignore it!

S + 7 = (3/4)(s + 7)    (3)

We can now solve equations (1), (2) and (3) for S, b and s.

Can you take it from here?

(You should get S = 8, b = 4 and s = 13).

Trapezoid HGFE is inscribed in a circle, with EF || GH. If arc GH is 70 degrees, arc EH is x^2 - 2x degrees and arc FG is 48 - 4x degrees, where x > 0 find arc EPF, in degrees.

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x² - 2x = 48 - 4x         x = 6

48 - 4 * 6 = 24

Arc   EPF = 360 - (70 + 2 * 24) = 242 degrees

Thanks Pangolin14  and Yugoslav     

I tackled this in a different way.

I also used co-ordinate geometry.  As shown on my pic.

I set the centre of the incircle (since that is what I is) as the origin.

B(-4,-3)  and  C((4,-3)    

I need to find the height of the triangle 

Let the equation of the line AB be     y= -mx+b  where m and b are both positive.

The perpendicular distance of the origin to line AB is the radius, 3.

So I will use the perpendicular distance formula

y= -mx+b

mx + y -b = 0        (0,0)

\(3=\frac{|0+0-b|}{\sqrt{m^2+1}}\\ 3=\frac{b}{\sqrt{m^2+1}}\\ b=3\sqrt{m^2+1} \)

So we have

\(y=-mx+3\sqrt{m^2+1}\\\)

This line passes through the point (4,-3) so substitute

\(-3=-4m+3\sqrt{m^2+1}\\ -3+4m=3\sqrt{m^2+1}\\ 16m^2-24m+9=9(m^2+1)\\ 16m^2-24m+9=9m^2+9\\ 7m^2-24m=0\\ 7m-24=0\\ m=\frac{24}{7}\)

\(b=3\sqrt{m^2+1}\\ b=3\sqrt{(\frac{24}{7})^2+1}\\ b=3*\frac{25}{7}=\frac{75}{7}\\ \text{height of triangle }= \frac{75}{7}+3 = \frac{96}{7}\)

\(\text{Area of triangle ABC = }\frac{1}{2}*8*\frac{96}{7}\\ \text{Area of triangle ABC = }54\frac{6}{7}\;units^2\)

Let ABCD be a rectangle with sides AB = 3 and BC = 5.  The rectangle is folded so that C ends up on AB with AC = 2 and BC = 1.  Find the area of quadrilateral PCMN.

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Information of folded rectangle: 

AB =3      AC = 2        BC = 1         CM + BM = 5

CM² - (5 - CM)² = 1           CM = 2.6        BM = 2.4

∠ACP ≅ ∠BMC   ==>         β = tan^-1(1/2.4)

DN = 2           AN = 3

[PCMN] = 6 ¹/₁₅

W+3=L

W(W+3)=70

Solve it as a quadratic.

A triangle has sides of 6, 8, and 10 inches.  What is the distance between the incenter and circumcenter of the triangle?

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Inradius:  r = 2     

Circumradius:  R = 5

The distance between the incenter and circumcenter is √5     

Let

\(f(t) = \int_0^t (x - 7)(x - 4)(x + 1)(e^x - 1) \left( \arctan (x) - \frac{\pi}{3} \right) \: dx \) .

Find all values of t where f(t) has a local minimum at t.

Hello Guest!

All values of t, where \(f(x) = (x - 7)(x - 4)(x + 1)(e^x - 1) \left( \arctan (x) - \frac{\pi}{3} \right) \: \)

has a local minimum at x ar, ar  \(x\in \{0,\ 5.712\}\)

 Determined graphically.            

  !

A+B=C

2B+C=AA=(6A+A)=7A

A+B-C=0

2B+C-7A=0

3B-6A

With trial and error we can find A=1 B=2 C=3

Use complementary counting and count the number of integers that don't contain 5 OR 6 and subtract it from the total numbers.

Ways to pick the first digit: 4. Possible ways: 1, 2, 3, 4

Second digit: 8 because you can't have 5 or 6

Third digit: 8 same as the second

4*8*8=256

Total numbers: 600

600-256=344

:D

You can use the principle of inclusion and exclusion. 1/3+2/7+1/4-(1/3*2/7)-(1/3*1/4)-(2/7*1/4)+(1/3*2/7*1/4)=9/14

:D

You could write out every term 3, 3/2, 3/4, 3/8, 3/16, 3/32, 3/64, 3/128

OR

Use the formula a*r^(n-1) where a is the first term, r is the ratio and n is the term you are solving for.

So 3*1/2^7 so you get 3/128

:D

Try making a venn diagram and use PIE

3/128

The denominator is two two the (x-1) power (x is the position in th sequence)

8-1=7

2^7=128

There are 195 people who have SOMETHING.

161+58+74= 293      So there is (293-195) = 98 overlap (double or triple counting)

people that have car and bike =a 

                            Car and pass = b

                             Pass and Bike =c      (Draw a 3 circle Venn Diagram to visualize)

a b and c are double counted    and    24  is triple counted

    So    a+b+c   + 2*24 = 98

         Then   a+b+c =50

Car only +24 +a   +b        + bike only +24 +a+c       + pass only + 24 + b + c   = 293

car only + bike only + pass only    + 2 (a+b+c)  + 3*24 =293

car only + bike only+ pass only = 293 - 2(50)-3(24)= 121