Dude!! If you want to post stupid sh** like this on here, then create an account. Otherwise just use this for math

It is 2or more die or one dice

Possible derivation:
d/dx(x^(2 cos(x)))
Express x^(2 cos(x)) as a power of e: x^(2 cos(x)) = e^(log(x^(2 cos(x)))) = e^(2 log(x) cos(x)):
  =  d/dx(e^((2 cos(x)) log(x)))
Using the chain rule, d/dx(e^(2 log(x) cos(x))) = ( de^u)/( du) ( du)/( dx), where u = 2 log(x) cos(x) and ( d)/( du)(e^u) = e^u:
  =  (d/dx(2 cos(x) log(x))) e^(2 cos(x) log(x))
Express e^(2 log(x) cos(x)) as a power of x: e^(2 log(x) cos(x)) = e^(log(x^(2 cos(x)))) = x^(2 cos(x)):
  =  x^(2 cos(x)) d/dx(2 cos(x) log(x))
Factor out constants:
  =  2 d/dx(cos(x) log(x)) x^(2 cos(x))
Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = cos(x) and v = log(x):
  =  log(x) d/dx(cos(x))+cos(x) d/dx(log(x)) 2 x^(2 cos(x))
The derivative of cos(x) is -sin(x):
  =  2 x^(2 cos(x)) (cos(x) (d/dx(log(x)))+-sin(x) log(x))
The derivative of log(x) is 1/x:
Answer: |  =  2 x^(2 cos(x)) (-(log(x) sin(x))+1/x cos(x))

14.7^2=216.09=216 9/100

Possible derivation:
d/dw(log(w)+(log(2, w))/(tan(w)))
Rewrite the expression: log(w)+(log(2, w))/(tan(w)) = log(w)+(cot(w) log(w))/(log(2)):
  =  d/dw(log(w)+(cot(w) log(w))/(log(2)))
Differentiate the sum term by term and factor out constants:
  =  d/dw(log(w))+(d/dw(cot(w) log(w)))/(log(2))
The derivative of log(w) is 1/w:
  =  (d/dw(cot(w) log(w)))/(log(2))+1/w
Use the product rule, d/dw(u v) = v ( du)/( dw)+u ( dv)/( dw), where u = cot(w) and v = log(w):
  =  1/w+log(w) (d/dw(cot(w)))+cot(w) (d/dw(log(w)))/(log(2))
The derivative of cot(w) is -csc^2(w):
  =  1/w+(cot(w) (d/dw(log(w)))+-csc^2(w) log(w))/(log(2))
The derivative of log(w) is 1/w:
  =  1/w+(-(csc^2(w) log(w))+1/w cot(w))/(log(2))
Simplify the expression:
Answer: | =  1/w+((cot(w))/w-csc^2(w) log(w))/(log(2))

There aren't any like terms or a GCF, so the expression is already simplified.

12n+4m-7

Solve the following system:
{-6 x-2 y = 20 |     (equation 1)
x-y = -6 |     (equation 2)
Add 1/6 × (equation 1) to equation 2:
{-(6 x)-2 y = 20 |     (equation 1)
0 x-(4 y)/3 = -8/3 |     (equation 2)
Divide equation 1 by 2:
{-(3 x)-y = 10 |     (equation 1)
0 x-(4 y)/3 = -8/3 |     (equation 2)
Multiply equation 2 by -3/4:
{-(3 x)-y = 10 |     (equation 1)
0 x+y = 2 |     (equation 2)
Add equation 2 to equation 1:
{-(3 x)+0 y = 12 |     (equation 1)
0 x+y = 2 |     (equation 2)
Divide equation 1 by -3:
{x+0 y = -4 |     (equation 1)
0 x+y = 2 |     (equation 2)
Collect results:
Answer: | {x = -4         and             y=2
 

It states his unpaid balance is                   3051.05

3051.05 x .37% = 3062.34   (rounded)

New Purchases  214.77 + 232.8 =              447.57     (these will accrue interest on the next billing cycle)

Payments                                                   -138.64

New Balance:                                            3371.26

Mike has a revolving credit account that has a monthly percentage rate of 0.37% on the unpaid balance. His unpaid balance for the July billing cycle is $3051.05. During the billing cycle, Mike purchased a stove for $214.77 and various tools for $232.80. Mike made a payment of $138.64. Find the new balance if the account uses the unpaid balance method. Round your solution to the nearest penny.

3051.05 + 214.77 + 232.80 - 138.64=3,359.98 Balance for "August" cycle.

3,359.98 X .37/100 =12.43 interest on unpaid balance

3,359.98 + 12.43 =$3,372.41 Principal and interest balance for August cycle.

What is the first term of the geometric sequence presented in the table below?
 

n      3      6

a_n  12   −324

\(\small{ \boxed{~ \text{geometric sequence: } \quad a_k = a_i^{ \frac{j-k}{j-i} }\cdot a_j^{ \frac{k-i}{j-i} } ~} } \)

\(\begin{array}{rcll} a_3 &=& a_i = 12 \qquad & i = 3\\ a_6 &=& a_j =-324 \qquad & j = 6\\ a_1 &=& a_k = \ ? \qquad & k = 1 \end{array}\)

\(\begin{array}{rcll} a_1 &=& 12 ^{ \left( \frac{6-1}{6-3} \right)} \cdot (-324)^{ \left( \frac{1-3}{6-3} \right) } \\ a_1 &=& 12 ^{\left( \frac{5}{3} \right) } \cdot (-324)^{ \left( \frac{-2}{3} \right) } \\ a_1 &=& \frac{ 12 ^{\left( \frac{5}{3} \right) } } { (-324)^{ \left( \frac{2}{3} \right) } }\\ a_1 &=& \sqrt[3]{ \frac{ 12^5 } { (-324)^{2} } } \\ a_1 &=& \sqrt[3]{ \frac{ 248832 } { 104976 } } \\ a_1 &=& \sqrt[3]{ 2.37037037037 } \\ a_1 &=& 1.33333333333 \\ a_1 &=& \frac43 \\ \end{array}\)

Simplify the following:
(2 (3 x+9))/(3)-(2 x+5)/(2)

Put each term in (2 (3 x+9))/3-(2 x+5)/(2) over the common denominator 6: (2 (3 x+9))/3-(2 x+5)/(2)  =  (4 (3 x+9))/6+(3 (-5-2 x))/6:
(4 (3 x+9))/6+(3 (-5-2 x))/6

(4 (3 x+9))/6+(3 (-5-2 x))/6 = (4 (3 x+9)+3 (-5-2 x))/6:
(4 (3 x+9)+3 (-5-2 x))/6

4 (3 x+9) = 12 x+36:
(12 x+36+3 (-5-2 x))/6

3 (-5-2 x) = -15-6 x:
(12 x+-15-6 x+36)/6

Grouping like terms, 12 x-6 x-15+36 = (12 x-6 x)+(36-15):
((12 x-6 x)+(36-15))/6

12 x-6 x = 6 x:
(6 x+(36-15))/6

36-15 = 21:
(6 x+21)/6

Factor 3 out of 6 x+21:3 (2 x+7)/6 3/6 = 3/(3×2) = 1/2: Answer: | | (2x+7)/2

I was understanding the general grasp of the question then it said round to the nearest penny. I thought I was dealing with dollars. Mind blown

\(2^{4-4-3-2-1}=2^{-6}=\frac{1}{64}=0,015625\)

h(t)=3500 ln(t+1)     we need to find the derivative here

h' (t)  = 3500 [ 1 / (t + 1) ]   =   3500/ [t + 1]

So

h ' (5)  =  3500 / [ 5 + 1]   ≈  583.3 ft/ min

Common mutiples of 2 and 3 up to 28

2  4  6  8  10  12  14  16  18 20  22  24  26  28

3  6  9  12  15  18  21  24  27

Now....just choose the numbers common to both lists =

6  12  18   24

And those are the common multiples

I'm assuming this is :

5^[(z+1)/3] = 25^z      if so, we can write

5^[ (z + 1) / 3 ]  = (5^2)^z

5^[ (z + 1)/ 3 ]   = 5^(2z)      since the bases are the same, we can solve for the exponents

[ z + 1 ] / 3   =  2z       multiply both sides by 3

z + 1  = 6z       subtract z  from both sides

1 = 5z             divide both sides by 5

1/5   = z

Solve for z over the real numbers:
5^(z+1)/3 = 25^z

Take the natural logarithm of both sides and use the identities log(a b) = log(a)+log(b) and log(a^b) = b log(a):
log(5) (z+1)-log(3) = 2 log(5) z

Expand out terms of the left hand side:
log(5) z-log(3)+log(5) = 2 log(5) z

Subtract 2 z log(5)-log(3)+log(5) from both sides:
-(log(5) z) = log(3)-log(5)

Divide both sides by -log(5):
Answer: |  z = 1-(log(3))/(log(5))=0.317393805......