All Answers 353127 Answers

hold the carrots bring the rabbit over drop the carrots and the bunny off the bunny wont run away because it is eating the carrots then bring the fox over and if it tryied arracking you then just kick it off

Take rabbit across

go back

Take fox across and go back across with rabbit

take carrots across and leave 'em with fox

go back and get rabbit....

Thanks Omi and CPhill, you guys are really helpful!

yes there you go

Next time, you should post your questions here in the "Questions Section" !!!

100 - 2x = 84      subtract 100 from both sides

- 2x = 84 - 100

-2x = - 16            divide both sides by - 2

x = - 16 / -2

x = 8 - Questions that Trevor answered incorrectly.

Rectangle ABCD is folded along BD, and point C lands on C1. BC1 and AD intersect at point E.

AB=5, AD=10.

What is the length of DE?

\(\text{Let $AB=DC=b$ } \\ \text{Let $AD=BC=a$ } \\ \text{Let $DE=x$ } \\ \text{Let $CC'=p$ } \\ \text{Let $EE'=y$ } \\ \text{Let $BD=d$ } \)

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{d} &\mathbf{=}& \mathbf{\sqrt{a^2+b^2}} \\ \hline & b^2 &=& p\cdot a \\ (2) & p &=& \dfrac{b^2}{a} \\ \hline & y^2 &=& p(p-a) \\ & &=& \dfrac{b^2}{a}\left(\dfrac{b^2}{a}-a \right) \\ & &=& \dfrac{b^2}{a}\left(\dfrac{a^2+b^2}{a} \right) \\ & &=& \dfrac{b^2}{a^2}\left(a^2+b^2\right) \\ (3) &\mathbf{y } &\mathbf{=}& \mathbf{\dfrac{b}{a}\sqrt{a^2+b^2}} \\ \hline & x^2 &=& \left(\dfrac{d}{2}\right)^2 + \left(\dfrac{y}{2}\right)^2 \\ & &=& \left(\dfrac{\sqrt{a^2+b^2}}{2}\right)^2 + \left(\dfrac{\dfrac{b}{a}\sqrt{a^2+b^2}}{2}\right)^2 \\ & &=& \dfrac{a^2+b^2}{4} + \dfrac{b^2}{a^2}\cdot \dfrac{(a^2+b^2)}{4} \\ & &=& \dfrac{a^2+b^2}{4} \left( 1+ \dfrac{b^2}{a^2}\right) \\ & &=& \dfrac{(a^2+b^2)^2}{4a^2} \\ &\mathbf{x } &\mathbf{=}& \mathbf{\dfrac{(a^2+b^2)}{2a}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a&=&10 \\ b &=& 5 \\\\ DE=x &=& \dfrac{5^2+10^2}{2\cdot 10} \\ x &=& \dfrac{125}{20} \\ x &=& \dfrac{25}{4} \\ \mathbf{x } &\mathbf{=}& \mathbf{6.25} \\ \hline \end{array}\)

The length of DE is 6.25

On a science test, 2 points are deducted from a total of 100 points for each question answered incorrectly. Trevor scored an 84 on the test. The equation 100 – 2x = 84, where x represents the number of questions answered incorrectly, represents the situation. How many questions did Trevor answer incorrectly?

 questions

Bob has 45 coins in her piggy bank. She has one fewer nickel than she has dimes, and one fewer quarter than three times the number of nickels. If Bob has only nickels, dimes, and quarters, how many nickels does she have?

N=number of nickels

D=number of dimes

Q=number of quarters

N+D+Q = 45

N=D-1

Q=3N+1

  = 3(D-1)-1

  = 3D-3-1

  = 3D - 4

so

N=D-1

Q=3D-4

D= D

D-1 + 3D - 4 + D = 45

                5D - 5 =  45

                      5D = 50

                        D = 10

N = 10-1 = 9

She has 9 nickels

Say you buy no steak at all then all the money goes on chicken 100/1.29 = 77.5 lbs of chicken

Say you buy no chicken and all the money goes on steak 100/3.49  = 28.6 pounds of steak.

I can graph this. You can purchase any amount of the 2 types of meat that lies inside the triangle

Help needed with the coordinates of points of intersection!

a:

\(\begin{array}{|rcll|} \hline y &=& x^2-6x \\ y &=& -8 \\\\ x^2-6x &=& -8 \\ x^2-6x+8 &=& 0 \\ x &=& \dfrac{6\pm \sqrt{36-4\cdot 8}}{2} \\ x &=& \dfrac{6\pm \sqrt{4}}{2} \\ x &=& \dfrac{6\pm 2}{2} \\ x &=& 3\pm 1 \\ \\ x_1 &=& 4 \qquad y_1 = -8 \\ x_2 &=& 2 \qquad y_2 = -8 \\ \hline \end{array} \)

b:

\(\begin{array}{|rcll|} \hline y&=& x^2-24 \\ y &=& -2x \\\\ x^2-24 &=& -2x \\ x^2+2x-24 &=& 0 \\ x &=& \dfrac{-2\pm \sqrt{4-4\cdot (-24)}}{2} \\ x &=& \dfrac{-2\pm \sqrt{100}}{2} \\ x &=& \dfrac{-2\pm 10}{2} \\ x &=& -1\pm 5 \\\\ x_1 &=& 4 \qquad y_1 = -2\cdot 4 = -8 \\ x_2 &=& -6 \qquad y_2 = -2\cdot (-6) = 12 \\ \hline \end{array}\)

The waterjet's form that spews up from the nozzle is drawn as the parabola. The formula is h= -0,9a^2 + 2,7a. 

A. 2,30 metres right of the nozzle Beatrice walks. Her height is 1,50 metres. Will she get wet if she stands there?

\(\begin{array}{|rcll|} \hline h(a) &=& -0,9a^2 + 2,7a \\ h(2,30) &=& -0,9\cdot 2,30^2 + 2,7\cdot 2,30 \\ h(2,30) &=& 1.449\ \text{metres.} \\ \hline \end{array}\)

She will get wet.

B. After what distance does the water come down on the ground again?

\(\begin{array}{|rcll|} \hline h(a) &=& -0,9a^2 + 2,7a \quad &| \quad h(a) = 0 \\ 0 &=& -0,9a^2 + 2,7a \\ a\cdot (-0,9a+2,7) &=& 0 \\\\ -0,9a+2,7 &=& 0 \\ a &=& \frac{2,7}{0,9} \\ a &=& 3 \ \text{metres.} \\ \hline \end{array} \)

After 3 metres does the water come down on the ground again.

C. Chris walks to the exact spot where the water is highest. He is 1,60 metres long. Will he get wet?

\(\text{The highest Point at $\dfrac{3 \ \text{metres.}}{2} = 1,5 $ metres. }\)

\(\begin{array}{|rcll|} \hline h(1,5) &=& -0,9\cdot 1,5^2 + 2,7\cdot 1,5 \\ h(1,5) &=& 2,025\ \text{metres.} \\ \hline \end{array}\)

He will not get wet.

answer is 523.3

answer is 28

In rectangle ABCD, P is the center point of DC. Q is a moving point on AB (doesn't overlap with A or B).

If the length of AQ is x and the sum of QM and QN is y,

choose the graph below that best describes the relationship between x and y.

\(\text{Let $\angle A =\angle\ PAB = \angle\ PBA $ } \\ \text{Let $AQ = x $ } \\ \text{Let $QM + QN = y $ } \)

\(\begin{array}{|rcl|rcl|} \hline \sin(A) &=& \dfrac{QM}{x} & \sin(A) &=& \dfrac{QN}{AB-x} \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline y(x) &=& QM + QN \\ &=& x \sin(A) + (AB-x) \sin(A) \\ &=& \sin(A) ( x + AB-x) \\ &=& AB\sin(A) \quad & \quad \text{y is constant} \\ \hline \end{array}\)

D.

f(x)   =  ceiling function  [ 1 / (x + 2) ]    x >-2

f(x)  =  floor function [ 1 / (x + 2) ]  x < -2

Notice that the first function can never  evaluate to 0  when x > -2

And the second function can never evaluate to 0  when x < - 2

So  0  is not in the range of f(x)

[ x^2 + 10x  + 21]

______________

[ x^2 + 4x  -  21 ]  

The domain excludes any x values that makes the denominator  = 0

So

x^2 + 4x  -  21  =  0

(x + 7) ( x - 3)  = 0

Setting each factor to 0 and solving for x  produces the two x values not in the domain  ⇒

x  =  -  7          and  x   = 3

So.....the domain  is

(-inf, -7) U (-7, 3)  U (3, inf)

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       

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

Note that, by AAS, triangle EC₁D   is congruent to triangle EAB

C₁ D  = AB  =  DC  = 5

BC₁  = BC  = 10

And BD  = √[C₁D^2 + BC₁^2]  = √ [5^2 + 10^2 ]  = √125

Note that  BE^2   = AE^2 + AB^2   =  AE^2 + 25

And  BE  = √[AE^2 + 25 ]

By the Law of Cosines, we have

AB^2  = AE^2 + BE^2  - 2 (AE * BE) cosAEB

5^2  = AE^2 + (AE^2 + 25) - 2(AE√[AE^2 + 25] )cos (AEB)

25  = AE^2 + AE^2 + 25 - 2 (AE √[AE^2 + 25] )cos(AEB)

[-2AE^2] / [ -2AE√[AE^2 + 25]   = cos(AEB)

AE/ √[AE^2 + 25]  = cos(AEB)    (1)

Note that  AEB  and DEB  are supplemental....so  cos(DEB)  = -cos(AEB)

So....using the Law of Cosines again, we have

BD^2  =  BE^2 + BE^2  - 2(BE^2)(-cos(AEB))

125 = 2BE^2 + 2BE^2(cos(AEB) )

125 = 2[AE^2 + 25]  - 2[AE^2 + 25] os(AEB)

125 = 2AE^2 + 50 - 2[AE^2 + 25] cos(AEB)

[75 - 2AE^2[ / [ 2(AE^2 + 25) ]  = cos(AEB)   (2)

Equate  (1)  and (2)

[75 - 2AE^2 ] / [ 2(AE^2 + 25)] = AE/ √[AE^2 + 25] 

[75 - 2AE^2] /(AE^2 + 25)  = 2AE/ √[AE^2 + 25] 

[75- 2AE^2] / (AE^2 + 25)  = 2AE√[AE^2 + 25] / (AE^2 + 25)

75 - 2AE^2  = 2AE√[AE^2 + 25]     square both sides

5625 - 300AE^2 + 4AE^4   = 4AE^2 [ AE^2 + 25]     simplify

5625 - 300AE^2 + 4AE^4 = 4AE^4 + 100AE^2 

5625 =  400AE^2

5626/400 = AE^2

225/16  = AE^2

15/4  = AE

So

DE  =  AD  - AE

DE  =  10  - 15/4

DE  = 40/4  - 15/4

DE  = 25/4   =  6.25

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)

came up with another go for it...no idea if it's correct.

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