Answer:
no it does not have to
Step-by-step explanation:
answer :
120 + 6 = 126
explanation:
sa calculator lang yan haha jk
pero yan talaga ang sagot promise hindi ako ng bo-bola haha stay safe and stay positive na lang huwag lang sa c o vi d
Answer:
x= -12
Step-by-step explanation:
Simplifying
4x + 10 = 2x + -14
Reorder the terms:
10 + 4x = 2x + -14
Reorder the terms:
10 + 4x = -14 + 2x
Solving
10 + 4x = -14 + 2x
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add '-2x' to each side of the equation.
10 + 4x + -2x = -14 + 2x + -2x
Combine like terms: 4x + -2x = 2x
10 + 2x = -14 + 2x + -2x
Combine like terms: 2x + -2x = 0
10 + 2x = -14 + 0
10 + 2x = -14
Add '-10' to each side of the equation.
10 + -10 + 2x = -14 + -10
Combine like terms: 10 + -10 = 0
0 + 2x = -14 + -10
2x = -14 + -10
Combine like terms: -14 + -10 = -24
2x = -24
Divide each side by '2'.
x = -12
Simplifying
x = -12
<u>Answer</u>
C. 4 + 6
D. 3.15151515.. + 4.6
E. √64 × √25
<u>Explanation</u>
A rational number is a number that can be expressed as a quotient or a fraction. That is it can be written as a/b.
π×6 = 6π ⇒ can not be written as a rational number.
√3 + √5 ⇒ the sum of two irrational numbers can not be rational.
4 + 6 = 10. Ten is a rational number. it can be written as 10/1
3.15151515.. + 4.6 = 104/33 + 23/5
= 1279/165
√64 × √25 = 8×5=40 This can be written as 40/1
Compute the definite integral:
integral_0^1 (5 x + 8)/(x^2 + 3 x + 2) dx
Rewrite the integrand (5 x + 8)/(x^2 + 3 x + 2) as (5 (2 x + 3))/(2 (x^2 + 3 x + 2)) + 1/(2 (x^2 + 3 x + 2)):
= integral_0^1 ((5 (2 x + 3))/(2 (x^2 + 3 x + 2)) + 1/(2 (x^2 + 3 x + 2))) dx
Integrate the sum term by term and factor out constants:
= 5/2 integral_0^1 (2 x + 3)/(x^2 + 3 x + 2) dx + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx
For the integrand (2 x + 3)/(x^2 + 3 x + 2), substitute u = x^2 + 3 x + 2 and du = (2 x + 3) dx.
This gives a new lower bound u = 2 + 3 0 + 0^2 = 2 and upper bound u = 2 + 3 1 + 1^2 = 6: = 5/2 integral_2^6 1/u du + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx
Apply the fundamental theorem of calculus.
The antiderivative of 1/u is log(u): = (5 log(u))/2 right bracketing bar _2^6 + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx
Evaluate the antiderivative at the limits and subtract.
(5 log(u))/2 right bracketing bar _2^6 = (5 log(6))/2 - (5 log(2))/2 = (5 log(3))/2: = (5 log(3))/2 + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx
For the integrand 1/(x^2 + 3 x + 2), complete the square:
= (5 log(3))/2 + 1/2 integral_0^1 1/((x + 3/2)^2 - 1/4) dx
For the integrand 1/((x + 3/2)^2 - 1/4), substitute s = x + 3/2 and ds = dx.
This gives a new lower bound s = 3/2 + 0 = 3/2 and upper bound s = 3/2 + 1 = 5/2: = (5 log(3))/2 + 1/2 integral_(3/2)^(5/2) 1/(s^2 - 1/4) ds
Factor -1/4 from the denominator:
= (5 log(3))/2 + 1/2 integral_(3/2)^(5/2) 4/(4 s^2 - 1) ds
Factor out constants:
= (5 log(3))/2 + 2 integral_(3/2)^(5/2) 1/(4 s^2 - 1) ds
Factor -1 from the denominator:
= (5 log(3))/2 - 2 integral_(3/2)^(5/2) 1/(1 - 4 s^2) ds
For the integrand 1/(1 - 4 s^2), substitute p = 2 s and dp = 2 ds.
This gives a new lower bound p = (2 3)/2 = 3 and upper bound p = (2 5)/2 = 5:
= (5 log(3))/2 - integral_3^5 1/(1 - p^2) dp
Apply the fundamental theorem of calculus.
The antiderivative of 1/(1 - p^2) is tanh^(-1)(p):
= (5 log(3))/2 + (-tanh^(-1)(p)) right bracketing bar _3^5
Evaluate the antiderivative at the limits and subtract. (-tanh^(-1)(p)) right bracketing bar _3^5 = (-tanh^(-1)(5)) - (-tanh^(-1)(3)) = tanh^(-1)(3) - tanh^(-1)(5):
= (5 log(3))/2 + tanh^(-1)(3) - tanh^(-1)(5)
Which is equal to:
Answer: = log(18)
