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Rus_ich [418]
3 years ago
9

Tom had 8 cookies, and he ate one and a half cookies. how many cookies were left?

Mathematics
1 answer:
Alex777 [14]3 years ago
6 0
Tom had 6 and a half cookies
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Margo won 24 out of her last 36 tennis matches.At this rate,predict how many matches she would win out of her next 27 matches ?
adell [148]
27 out of 39 win i guess If u think this is right ok
7 0
4 years ago
Please help solve for PQ.
DerKrebs [107]
This problem cannot be solved unless we are given figure NPQR is a rhombus.
In that case, then all sides are equal, meaning 
5x+16=9x-32
Solve for x
9x-5x=16+32
4x=48
x=12

Each side (including PQ) then equals 5x+16=5*12+16=76
8 0
3 years ago
What's the difference between (-3)^2 and -3^2
erastova [34]
(-3)^2 = 9
-3^2 = -9
Hope this helps!
8 0
4 years ago
Read 2 more answers
Solving Rational Functions Hello I'm posting again because I really need help on this any help is appreciated!!​
Greeley [361]

Answer:

x = √17 and x = -√17

Step-by-step explanation:

We have the equation:

\frac{3}{x + 4}  - \frac{1}{x + 3}  = \frac{x + 9}{(x^2 + 7x + 12)}

To solve this we need to remove the denominators.

Then we can first multiply both sides by (x + 4) to get:

\frac{3*(x + 4)}{x + 4}  - \frac{(x + 4)}{x + 3}  = \frac{(x + 9)*(x + 4)}{(x^2 + 7x + 12)}

3  - \frac{(x + 4)}{x + 3}  = \frac{(x + 9)*(x + 4)}{(x^2 + 7x + 12)}

Now we can multiply both sides by (x + 3)

3*(x + 3)  - \frac{(x + 4)*(x+3)}{x + 3}  = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}

3*(x + 3)  - (x + 4)  = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}

(2*x + 5)  = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}

Now we can multiply both sides by (x^2 + 7*x + 12)

(2*x + 5)*(x^2 + 7x + 12)  = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}*(x^2 + 7x + 12)

(2*x + 5)*(x^2 + 7x + 12)  = (x + 9)*(x + 4)*(x+3)

Now we need to solve this:

we will get

2*x^3 + 19*x^2 + 59*x + 60 =  (x^2 + 13*x + 3)*(x + 3)

2*x^3 + 19*x^2 + 59*x + 60 =  x^3 + 16*x^2 + 42*x + 9

Then we get:

2*x^3 + 19*x^2 + 59*x + 60 - (  x^3 + 16*x^2 + 42*x + 9) = 0

x^3 + 3x^2 + 17*x + 51 = 0

So now we only need to solve this.

We can see that the constant is 51.

Then one root will be a factor of 51.

The factors of -51 are:

-3 and -17

Let's try -3

p( -3) = (-3)^3 + 3*(-3)^2 + +17*(-3) + 51 = 0

Then x = -3 is one solution of the equation.

But if we look at the original equation, x = -3 will lead to a zero in one denominator, then this solution can be ignored.

This means that we can take a factor (x + 3) out, so we can rewrite our equation as:

x^3 + 3x^2 + 17*x + 51 = (x + 3)*(x^2 + 17) = 0

The other two solutions are when the other term is equal to zero.

Then the other two solutions are given by:

x = ±√17

And neither of these have problems in the denominators, so we can conclude that the solutions are:

x = √17 and x = -√17

6 0
3 years ago
Given line BD is congruent to BC, find the measure of angle B
Degger [83]
If BD is congruent to BC, that means that the sides are equal, so their angles are too.
6x-9 = 3x+24
3x = 33
x = 33/3
x = 11
Angle BCD:
6×11-9 = 66-9 =57°
Angle BDC:
3×11+24 = 33+24 = 57°
Angle B:
x = 180° - 2×57°
x = 66°
5 0
3 years ago
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