hectictar

Let  AE  =  b

Let  DE  =  c

AE + DE  =  AD

b + c  =  10

And by the Pythagorean theorem....

b² + 5²  =  c²

b² + 25  =  c²

Now we can find  c .

b + c  =  10        so        b  =  10 - c     Use this value for  b  in the second equation.

b² + 25  =  c²

(10 - c)² + 25  =  c²

100 - 20c + c² + 25  =  c²

125 - 20c + c²  =  c²

125 - 20c  =  0

125  =  20c

125 / 20  =  c

6.25  =  c       

\(\begin{array}\ &&\frac{(a+b)^n}{b^n} \\ =&&\big(\frac{a+b}{b}\big)^n \\ =&&\big(\frac{b+a}{b}\big)^n \\ =&&\big(\frac{b}{b}+\frac{a}{b}\big)^n \\ =&&\big(1+\frac{a}{b}\big)^n \\ \end{array}\)

a.

x² + 4x  =  0       Factor an  x  out of the terms on the left side.

x(x + 4)  =  0       Set each factor equal to zero.

x = 0     or     x + 4 = 0

                     x  =  -4

The solutions are   x = 0   and   x = -4  .

b.

It means that when  x = 0  and when  x = -4 ,  y = 0 .   The x-intercepts are  0  and  -4 .

c.

y  =  x² + 4x

y  =  5

Since  y = 5  ,  we can plug in  5  for  y  in the first equation.

5  =  x² + 4x               Subtract  5  from both sides.

0  =  x² + 4x - 5          Factor the right side.

0  =  (x + 5)(x - 1)       Set each factor equal to zero.

x + 5  =  0        or        x - 1  =  0

x  =  -5             or        x  =  1

So the points of intersection for  A  and  B  are:   (-5, 5)   and   (1, 5)

d.

y  =  x² + 4x

y  =  2x

Since  y = 2x , we can plug in  2x  for  y  in the first equation.

2x  =  x² + 4x          Subtract  2x  from both sides.

0  =  x² + 4x - 2x

0  =  x² + 2x            Factor the right side.

0  =  x(x + 2)           Set each factor equal to zero.

x  =  0        or        x + 2 - 0

                             x  =  -2

When  x = 0 ,   y  =  2(0)  =  0     So one of the intersection points is   (0, 0)

When  x = -2 ,  y  =  2(-2)  =  -4     So the other intersection point is   (-2, -4)

The points of intersection for  A  and  C  are:  (0, 0)   and   (-2, -4)

a.   Solve the equation  x² - 2x  =  24

x² - 2x  =  24

                             Subtract  24  from both sides of the equation.

x² - 2x - 24  =  0

                             Factor the left side. What two numbers add to -2 and multiply to -24 ?  -6 and +4

(x - 6)(x + 4)  =  0

                             Set each factor equal to zero.

x - 6  =  0        or         x + 4  =  0

x  =  6             or         x  =  -4

b.   It means...

The two possible values of  x  that make the area of the triangle be  24  square cm  are  6  and  -4  .

However, since  x  is a length, we can exclude the negative possibility.

So  x  must be  6 .

When  x  is  6 ,  the area of the triangle is  24  square cm.

the height of the triangle   =   (x - 2) cm   =   (6 - 2) cm   =   4 cm

c.

area of the triangle  =  x² - 2x

                                                   Plug in  120  for the area of the triangle.

120  =  x² - 2x

                                                   Subtract  120  from both sides of the equation.

0  =  x² - 2x - 120

                                                   Factor the right side of the equation.

0  =  (x - 12)(x + 10)

                                                   Set each factor equal to zero.

x - 12  =  0        or        x + 10  =  0

x  =  12             or        x  =  -10

Exclude the negative possibility.

x  =  12

length of  AB   =   2 * x   =   2 * 12   =   24   cm

a

r  =  a² + 12a       Factor an  a  out of both terms on the right side.

r  =  a(a + 12)

b

s  =  b² - 8b + 7

s  =  b² - 8b + 7       What two numbers that add to  -8  and multiply to  7 ?    -7  and  -1

                                So the right side factors like this...

s  =  (b - 7)(b - 1)

c

t  =  c² - c - 30

t  =  c² - 1c - 30          What two numbers add to  -1  and multiply to  -30 ?   +5  and  -6  . So....

t  =  (c + 5)(c - 6)

d

u  =  d² + 6d - 7

u  =  d² + 6d - 7        What two numbers add to  6  and multiply to  -7 ?   +7  and  -1  . So...

u  =  (d + 7)(d - 1)

I also get  864  cm³  (but exactly) 

area of square base  =  288    so   side length of base  =  √288

Let  x  be the diagonal of the base.

Let  h  be the height of the pyramid.

By the Pythagorean theorem....

(√288)² + (√288)²   =   x²

288 + 288   =   x²

576   =   x²

x   =   √576

x   =   24

Again by the Pythagorean theorem...

(x/2)² + h²  =  15²

(24/2)² + h²  =  15²

144 + h²  =  225

h²  =  81

h   =   9

volume of pyramid   =   (1/3)(area of base)(height)

volume of pyramid   =   (1/3)(288)(9)

volume of pyramid   =   864      cm³

Let each small cube be  1 cubic unit.

volume of big cube  =  125 cubic units

side length of big cube  =  125^(1/3) units  =  5  units

surface area of big cube  =  total painted area  =  6 * 5^2  =  150  sq units

surface area of one small cube  =  6  sq units

combined surface area of small cubes  =  125 * 6  sq units  =  750 sq units

Each face of small cube is 1 sq unit.

There are a total of 150 sq units that are painted, so there are a total of 150 faces with paint on them.

There are a total of 750 sq units, so there are a total of 750 faces.

The probability that a randomly selected face is one with paint on it  =  150/750  =  1/5

f(x)  =  7x + 5

g(x)  =  x - 1

h(x)  =  f( g(x) )  =  f( x - 1 )  =  7(x - 1) + 5  =  7x - 7 + 5  =  7x - 2

h(x)  =  7x - 2

                            To find the inverse, let's replace  h(x)  with  y .

y  =  7x - 2

                            Solve for  x  ....add  2  to both sides of the equation.

y + 2  =  7x

                            Divide both sides by  7 .

(y + 2)/7  =  x

x  =  (y + 2)/7      So the inverse of  h(x)  is..

h^-1(x)  =  (x + 2)/7

These problems easily confuse me but here's my attempt....

surface area of N's donut holes   =   4 pi * (6 mm)²   =   144pi   mm²

surface area of T's donut holes   =   4 pi * (8 mm)²   =   256pi   mm²

surface area of A's donut holes   =   4 pi * (10 mm)²   =   400pi   mm²

When N finishes  1  donut, T will have finished  \(\frac{144}{256}\)  donuts and A will have finished  \(\frac{144}{400}\)  donuts.

When N finishes  1  donut, T will have finished  \(\frac{9}{16}\)  donuts and A will have finished  \(\frac{9}{25}\)  donuts.

When N finishes  2  donuts, T will have finished  2*\(\frac{9}{16}\)  donuts and A will have finished  2*\(\frac{9}{25}\)  donuts.

When N finishes  2  donuts, T will have finished  \(\frac98\)  donuts and A will have finished  \(\frac{18}{25}\)  donuts.

When N finishes  x  donuts, T will have finished  \(\frac{9x}{16}\)  donuts and A will have finished  \(\frac{9x}{25}\)  donuts.

We need the smallest integer  x  such that  \(\frac{9x}{16}\)  is an integer and  \(\frac{9x}{25}\)  is an integer.

For a fraction to be an integer, all the factors in the denominator must be in the numerator.

For  9x/16  to be an integer,  2*2*2*2  needs to be factors of  x  .

For  9x/25  to be an integer,  5*5  needs to be factors of  x .

So  x  must be  2*2*2*2*5*5   =   400

When Niraek finishes 400 donut holes, Theo will have finished  400*9/16 = 225  donut holes, and Akshaj will have finished  400*9/25 = 144  donut holes.