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telo118 [61]
3 years ago
12

Write 2lnx−3lny+1/2lnz as a single logarithm.

Mathematics
1 answer:
Alex3 years ago
3 0
2lnx−3lny+1/2lnz=lnx²-lny³+ln√z=(lnx²+ln√z)-lny³=ln(x²√z)-lny³=ln\frac{ x^{2}  \sqrt{z} }{ y^{3} }


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Every 23 hour, Harris can sew 16 pair of jeans.
Setler79 [48]

Answer:

16 : 23

Step-by-step explanation:

23 is a prime number, so it can't be reduced

6 0
3 years ago
Solve: VII multiplied by IX. Show your answer in standard form.
matrenka [14]
7*9
b. 63


v=5 i=1 
5+1+1= 7

x=10

10-1= 9

9*7= 63

8 0
3 years ago
There are 450 people at a play, the total receipt for $600 and the admission was $2 for adults and $0.75 for children, how many
sdas [7]

Answer:

340 children, 210 adults.

Step-by-step explanation:

Let \ a \ - adults, \ c - children;\\c + a = 450\\0.75c + 2a = 600|\cdot 4\\3c + 8a = 2400;\\\\c + a = 450|\cdot(-3)\\3c + 8a = 2400\\-3c -3a = -1350\\5a = 1050 => a = 210, c = 450 -210 = 340.

8 0
3 years ago
Use the Fundamental Theorem for Line Integrals to find Z C y cos(xy)dx + (x cos(xy) − zeyz)dy − yeyzdz, where C is the curve giv
Harrizon [31]

Answer:

The Line integral is π/2.

Step-by-step explanation:

We have to find a funtion f such that its gradient is (ycos(xy), x(cos(xy)-ze^(yz), -ye^(yz)). In other words:

f_x = ycos(xy)

f_y = xcos(xy) - ze^{yz}

f_z = -ye^{yz}

we can find the value of f using integration over each separate, variable. For example, if we integrate ycos(x,y) over the x variable (assuming y and z as constants), we should obtain any function like f plus a function h(y,z). We will use the substitution method. We call u(x) = xy. The derivate of u (in respect to x) is y, hence

\int{ycos(xy)} \, dx = \int cos(u) \, du = sen(u) + C = sen(xy) + C(y,z)  

(Remember that c is treated like a constant just for the x-variable).

This means that f(x,y,z) = sen(x,y)+C(y,z). The derivate of f respect to the y-variable is xcos(xy) + d/dy (C(y,z)) = xcos(x,y) - ye^{yz}. Then, the derivate of C respect to y is -ze^{yz}. To obtain C, we can integrate that expression over the y-variable using again the substitution method, this time calling u(y) = yz, and du = zdy.

\int {-ye^{yz}} \, dy = \int {-e^{u} \, dy} = -e^u +K = -e^{yz} + K(z)

Where, again, the constant of integration depends on Z.

As a result,

f(x,y,z) = cos(xy) - e^{yz} + K(z)

if we derivate f over z, we obtain

f_z(x,y,z) = -ye^{yz} + d/dz K(z)

That should be equal to -ye^(yz), hence the derivate of K(z) is 0 and, as a consecuence, K can be any constant. We can take K = 0. We obtain, therefore, that f(x,y,z) = cos(xy) - e^(yz)

The endpoints of the curve are r(0) = (0,0,1) and r(1) = (1,π/2,0). FOr the Fundamental Theorem for Line integrals, the integral of the gradient of f over C is f(c(1)) - f(c(0)) = f((0,0,1)) - f((1,π/2,0)) = (cos(0)-0e^(0))-(cos(π/2)-π/2e⁰) = 0-(-π/2) = π/2.

3 0
4 years ago
Would someone Please answer this question please will be thanked and also will pick brainly!! (please be honest)
lapo4ka [179]
The answer to this question is 360ft^2
8 0
4 years ago
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