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If you roll a 1, 2, 3, or 4, you earn $3. If you roll a 5 or 6, you earn $6. So, we can write the following:

$E = \frac{4}{6}(\$3) + \frac{2}{6}(\$6) = \boxed{\$4}$

Simplify the expressions for the line and circle:

y = (4/3)x - (10/3)

(x-1)^2 + (y+2)^2 = 25

Graphing both of these equations, we can see that the graphs intersect at (4, 2) and (-2, -6).

ElectricPavlov nice solution!

Solution for 2:

Since the perimeter of the square is $2\sqrt{2}$, the side length of one side of the square is $\frac{\sqrt{2}}{2}$. The diagonal of that square is then equal to $\frac{\sqrt{2}}{2} \cdot \sqrt{2} = 1$

Let x be the radius of one circle, and let y be the radius of the other circle. 

Notice that the diagonal is equal to $x\sqrt{2} + x + y\sqrt{2} + y$. Therefore, 

$x\sqrt{2} + x + y\sqrt{2} + y = 1\\ x(\sqrt{2}+1)+ y(\sqrt{2}+1)=1\\ x+y=\frac{1}{\sqrt{2}+1} = \sqrt{2}-1$

Since the sum of the circumference of the circle is 2 times the radius times pi, just multiply 2pi to get the final answer:

$\boxed{2\sqrt{2}\pi-2\pi}$

We can use the triangle inequality. The sum of two sides must be greater than the third:

7 + 12 > x

7 + x > 12

12 + x > 7

Solving, we get:

19 > x

x > 5

x > -5

We get the inequality 5 < x < 19, so there are 12 possible values of x: 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18.

1. A(2) - A(1) = 11 - 2 = 9.

2. The solution is (-inf,-4) U (3,inf).

3. The solultion is [1/4,3/2].

Triangle side length rule

7 + x     must be greater than 12          x >5

Let's rearrange the equation:$z^2 - 12z + 7 = 0$

This is a quadratic, so we can use the quadratic formula:

$x = {12 \pm \sqrt{144-4 \cdot 1 \cdot (-7)} \over 2(1)} = \frac{12 \pm 2\sqrt{43}}{2} = 6 \pm \sqrt{43}$

 a      b      and    1000-a-b    are the amounts   

2a   + 2b   + (1000-a-b -100)  = 1500

a + b  + 900 = 1500

a + b = 600

1000 - (a+b)  =   400    is what Al had  

a = k / b

-3 = k / -6

k = proportionality constant = 18

a = 18/b

a = 18/8 = 2 1/4     or   9/4

7x^2 + 5x - h = 0      for only ONE solution, the discriminant must = 0

b^2 - 4ac  = 0

5^2 - 4(7)(-h)

25 + 28 h = 0

h = -25/28

The numerator does not matter in this case; only if the denominator is zero for some real number, the domain will not be set for all real numbers.

To prevent the denominator from having real roots, the discriminant must be negative, which means $2^2-4\cdot-5\cdot k = 4+20k$ is negative.

That will only happen if $\boxed{k<-\frac{1}{5}}$.

We can plug in all the answer choices and see if it satisfies all the inequalities.

A) x = 4, y = 0

4 > 2, 0 = 0  (doesn't work)

B) x = 3, y = 1

3 > 2, 1 > 0, 3 + 1 < 5 (works)

C) x = 3, y = 2

3 > 2, 2 > 0, 3 + 2 = 5 (doesn't work)

D) x = 1, y = 3

1 < 2 (doesn't work)

E) x = 2, y = 2

2 = 2 (doesn't work)

So, the answer is B.

We can write this as a system of equations. Let c be the number of children and a be the number of adults who attended.

900 people were present at the game ==> a + c = 900      .  (1)

$6500 was collected ==> 9a + 5c = 6500        . (2)

We can use the elimination method to find a and c. Multiply equation (1) by 5:

5a + 5c = 4500

9a + 5c = 6500

Subtracting both equations yields -4a = -2000, so a = 500 and c = 400.

Apologies for the formatting, I am new to the site.

Ways to roll a sum of '6'  with two rolls

1 5

5 1

2 4

4 2

3 3       1 out of 5       =    1/5

Wow! Thank you for answering very fast.

Here is  A  way to solve first one....may not be THE BEST way....

     circle radius = sqrt 60

draw a line from origin to corne of square that touches circle     this line forms a   arc tan (1/3) = 18.435  degree angle

     sqrt 60 cos 18.435   = length of 3 boxes = 7.34      so each square is side length   7.34/3

              then total area of  1 square   =     (7.34/3  *  7.34/3) =  6 in²    

On a Cartesian coordinate plane, points $(1,2)$ and $(7, 4)$ are opposite vertices of a square. What is the area of the square?

Hello wizzymath!

$A=(\dfrac{d}{\sqrt{2}})^2=\dfrac{d^2}{2}$

$d^2=(x_2-x_1)^2+(y_2-y_1)^2\\ d^2=(7-1)^2+(4-2)^2\\ d^2=40$

$A=\dfrac{d^2}{2}=\dfrac{40}{2}$

$A=20$

The area of the square is 20.

  !
 

All of the equations in a system true.

Sure those will intersect....but lines of DIFFERENT slopes....whether positve OR negative will intersect

 +   +   will intersect if different     -   -  will intersect if different       +   -    will interesect

Perhaps the THIRD   fifth and ninth would be a r = 2

It takes four painters working at the same rate 1.25 work-days to finish a job. If only three painters are available, how many work-days will it take them to finish the job, working at the same rate?

Hello wizzymath!

$W=N\cdot t\cdot n\\ N=\dfrac{W}{t\cdot n}=\dfrac{W}{1.25d\cdot4}$

$W=\dfrac{W}{1.25d\cdot4}\cdot t\cdot 3\\ t=\dfrac{W\cdot 1.25d\cdot 4}{3\cdot W}$

$t=1\frac{2}{3}\ d$

Three painters need $1\frac{2}{3}$ working days for the same work.

x > 2      Answer A B and C

y>0                         B and C

x+y<5       Answer B

x = salary before increase

1.16 x = 55680

x = 55680/(1.16) =  $ 48000.

Ursala  Andre(ss) .... I wonder if that was a joke?

Anyway .    Andre(ss) has a 1 minute head start    =   48 meters ahead

 Ursala swims 60-48 = 12 meters/min  faster than Andre(ss) 

      Ursala will close that  48 -30 = 18  meters in   18 m / 12 m/s = 1  1/2  minutes to be 30 m behind Andre(ss)

Let the middle number be x.

The sum of the 3 numbers will be 3x+4.

Write the following equation : 

3x+4 = 67

Thus, 3x = 63,

And x = 21 

Thus, the largest number is 28 (21+7)