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   root 1     -7 + 2i       means   (x- (-7+2i))   is a factor  of the quadratic:   (x+7-2i)

          2                                    (x- (7 - 2i)    is a factor                               (x - 7+2i)

   ( I suspect the second root should be -7 - 2i)

(x+7-2i)(x-7+2i)      will be your quadratic ....but it will not be of the form    z^2 + bz + c = 0    

    if the second root is   (-2i - 7)    you will get the factor    (x+7+2i)

               then factor1 * factor2    would be      (x+7-2i)(x+7+2i)     and when multiplied out would be of the correct form to calculate b+c.....

How many are there altogether?

Which ones are prime?

Which ones are even?

Answer those and get back to me.

What have you done already, by yourself?:

hint:

pythagoras's theorum

then

similar triangles

I have responded.

There have been many questions recently very similar to this.

This is one of the easier ones.

Let D be the amount David starts with and E be the amount that Elijiah starts with.   You want to find   D:E

If David has a quarters of D left  he must have spent  (3/4)D

If Elijah has 3/5E left he must have spent  (2/5)E

They both spent the same amount so you can take it from there.

Answered

Kent, Mike and Zam shared 575 cards. The ratio of the number of cards Kent had to the number of cards Mike had was of 7:3. The ratio of the number of cards Mike had to the number of cards Zam had was 6:5. How many more cards did Kent have than Zam?

K:M = 7:3            M:Z = 6:5 = 3:2.5

K:M:Z = 7:3:2.5   = 14:6:5

575/(14+6+5) = 23

K = 6 * 23 =138

Z = 5 * 23 = 115

The answer is 138 - 115 

Answered  

A crate contained forks and knives. 3/5 of the cutlery in the crate were forks. 30 forks and 4 knives were removed from the crate and the remaining cutlery were divided into groups of 9 pieces each. In each group, there were 5 forks. What is the total number of cutlery in the crate at first?

Let there by F forks and K knives and a total of T pieces of cutlery

_{"A crate contained forks and knives. 3/5 of the cutlery in the crate were forks."}

so     F= 0.6T   and     K=0.4T

_{30 forks and 4 knives were removed from the crate}

So the number of forks left is  0.6T - 30

and the number of knives left is   0.4T - 4

Now there are divided into groups of 9  Every group has 5 forks and therefore every group must have 4 knives.

The ratio of forks to knives (that are remaining) must be  5:4

so

$\frac{0.6T-30}{0.4T-4}=\frac{5}{4}\\ 2.4T-120=2T-20\\ 0.4T=100\\ T=250$ 

Well 

G=M+13

and

G+M=113

now solve simultaneously.

The answer is 22 integers.

The probability is 7/10.

Paul has 5 quarters.

The smallest integer is 10.

The answer is 22.

Explanation:

There are 36 possible outcomes in rolling two six-sided cubes.

Of those 36 possibilities, five of them result in a sum of 6.

1+5:   2+4:   3+3:   4+2:   5+1

(1+5) is different from (5+1) use two different colors of dice such as black and white to make this obvious)

5 = the number of possibilities of getting a six.

36 = the total number of possibilities (6×6=36)

So the probability is 5/36

Thanks for taking time and helping me with this problem, and even including steps! I understand how to do it now! Thanks again!

Paul has a total of 23 quarters.

Explanation: First, add 5 to each side of the inequality to isolate the absolute value function while keeping the inequality balanced: | 2 x + 1 | − 5 + 5 < 0 + 5 | 2 x + 1 | − 0 < 5 | 2 x + 1 | < 5.

The absolute value function takes any term and transforms it to its non-negative form.

Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

− 5 < 2 x + 1 < 5.

Next, subtract 1 from each segment of the system of inequalities to isolate the x term while keeping the system balanced:

− 5 − 1 < 2 x + 1 − 1 < 5 − 1 − 6 < 2 x + 0 < 4 − 6 < 2 x < 4

Now, divide each segment of the system by 2 to solve for x while keeping the system balanced:

− 6 2 < 2 x 2 < 4 2 − 3 < 2 x 2 < 2 − 3 < x < 2 Or x > − 3 and x < 2

Or, in interval notation: ( − 3 , 2 )

By telescoping product, $P = 2^{5/3} = \sqrt[3]{32}$, so $a + b = 32 + 3 = \boxed{35}$.

The largest number of elements in S is 20, one for each congruence class modulo 5.

Jack has twice as many dimes as quaters.

Total value = $6.30

Total value = $6.30 = 630 cents

Now Let the total number of dimes be x, and total number of quaters be y.

Hence it is given,

                         
                            $x=2y$
 

Further we know that the total amount is 630 cents.

Therefore we can write it as,

                       $10x + 25y=630$
Now we shall solve the two equations using substitution, put ^{$x = 2y$} in equation $10x + 25y = 630$

we get, 

10 (2y) + 25y = 630

   20y  +  25y = 630

               45y = 630

                   y = 14

Now we know ^{$x = 2y,$} Therfore $x = 2(14)=28$

The total number of dimes he has are 28.

Is there some more information?

Maybe an equal sign?

=^._.^=