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
You're answer is D.
-3/-3(2)-9
1(2)-9
2-9
7
so it is D
The range of the <em>quadratic</em> function y = (2 / 3) · x² - 6 is {- 6, 0, 18, 48}.
<h3>What is the range of a quadratic equation?</h3>
In this case we have a <em>quadratic</em> equation whose domain is stated. The domain of a function is the set of x-values associated to only an element of the range of the function, that is, the set of y-values of the function. We proceed to evaluate the function at each element of the domain and check if the results are in the choices available.
x = - 9
y = (2 / 3) · (- 9)² - 6
y = 48
x = - 6
y = (2 / 3) · (- 6)² - 6
y = 18
x = - 3
y = (2 / 3) · (- 3)² - 6
y = 0
x = 0
y = (2 / 3) · 0² - 6
y = - 6
x = 3
y = (2 / 3) · 3² - 6
y = 0
x = 6
y = (2 / 3) · 6² - 6
y = 18
x = 9
y = (2 / 3) · 9² - 6
y = 48
The range of the <em>quadratic</em> function y = (2 / 3) · x² - 6 is {- 6, 0, 18, 48}.
To learn more on functions: brainly.com/question/12431044
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