Answer:
The values of p in the equation are 0 and 6
Step-by-step explanation:
First, you have to make the denominators the same. to do that, first factor 2p^2-7p-4 = \left(2p+1\right)\left(p-4\right)2p
2
−7p−4=(2p+1)(p−4)
So then the equation looks like:
\frac{p}{2p+1}-\frac{2p^2+5}{(2p+1)(p-4)}=-\frac{5}{p-4}
2p+1
p
−
(2p+1)(p−4)
2p
2
+5
=−
p−4
5
To make the denominators equal, multiply 2p+1 with p-4 and p-4 with 2p+1:
\frac{p^2-4p}{(2p+1)(p-4)}-\frac{2p^2+5}{(2p+1)(p-4)}=-\frac{10p+5}{(p-4)(2p+1)}
(2p+1)(p−4)
p
2
−4p
−
(2p+1)(p−4)
2p
2
+5
=−
(p−4)(2p+1)
10p+5
Since, this has an equal sign we 'get rid of' or 'forget' the denominator and only solve the numerator.
(p^2-4p)-(2p^2+5)=-(10p+5)(p
2
−4p)−(2p
2
+5)=−(10p+5)
Now, solve like a normal equation. Solve (p^2-4p)-(2p^2+5)(p
2
−4p)−(2p
2
+5) first:
(p^2-4p)-(2p^2+5)=-p^2-4p-5(p
2
−4p)−(2p
2
+5)=−p
2
−4p−5
-p^2-4p-5=-10p+5−p
2
−4p−5=−10p+5
Combine like terms:
-p^2-4p+0=-10p−p
2
−4p+0=−10p
-p^2+6p=0−p
2
+6p=0
Factor:
p=0, p=6p
Answer:
a set of two or more equations, each containing two or more variables whose values can simultaneously satisfy both or all the equations in the set, the number of variables being equal to or less than the number of equations in the set.
Step-by-step explanation:
It works because of two properties of equations: Multiplying (or dividing) the expression on each side by the same number does not alter the equation. Adding two equations produces another valid equation: e.g. 2x = x + 10 (x = 10) and x − 3 = 7 (x also = 10).
I'll do 22 for you. 22, 24, 26, 28, and 30, are very similar to each other.
PROPORTIONS:
Draw it out.
5/7 = either c/6, or b/5.
I'll do c/6
5/7 = c/6
solve for c
7c=30
c=30/7
now do the same thing to the remaining side, b.
there!
Step-by-step explanation:
Multiply the radius by 2 to get the diameter. Multiply the result by π, or 3.14 for an estimation. That's it; you found the circumference of the circle
And then for diameter you have to Multiply the diameter by π, or 3.14. The result is the circle's circumference.
