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
It would be : -9.3 , -4.125 and -2.25.
Hope This Helps You!
Good Luck Studying :)
Answer:
So I would say your answer is..............A
Step-by-step explanation:
For most places, global warming will result in more frequent hot days and fewer cool days, with the greatest warming occurring over land. Longer, more intense heat waves will become more common. Storms, floods, and droughts will generally be more severe as precipitation patterns change
Also, A is correct because lets say you put an ice cube in the sun...okay well if it keeps getting hotter and hotter multiply ice cubes into ice caps. Well same thing will happen. It will melt causeing water from the ice caps. Which will result into more water. Which will cause water to go above sea level!
Hope this helps!!!
Since
, we can rewrite the integral as

Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:

Both integrals are quite immediate: you only need to use the power rule

to get
![\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E11-3t%5E2%5C%3Bdt%20%3D%20%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%2C%5Cquad%20%5Cint_1%5E4%202t%5C%3B%20dt%20%3D%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4)
Now we only need to evaluate the antiderivatives:
![\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15](https://tex.z-dn.net/?f=%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%20%3D%201-1%5E3%3D0%2C%5Cquad%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4%20%3D%204%5E2-1%5E2%3D15)
So, the final answer is 15.