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Evgen [1.6K]
3 years ago
5

Which is the logarithmic form of 2^10=1024

Mathematics
1 answer:
vfiekz [6]3 years ago
6 0

Answer:

Convert the exponential equation to a logarithmic equation using the logarithm base

2^10=1024

10

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Solve the inequality using the algebraic approach: x+2>3x-8
kow [346]
X + 2 > 3x - 8
subtract x from both sides

2 > 2x - 8
add 8 to both sides

10 > 2x
divide both sides by 2

5 > x


ANSWER: 5 > x

Hope this helps! :)
4 0
4 years ago
Find the coordinates of the midpoint
love history [14]

Answer:

(2,2)

Step-by-step explanation:

(5+-1/2,8+-4/2)

(-2,2)

7 0
3 years ago
Someone help me please
Mazyrski [523]

Answer:

x = -7

Step-by-step explanation:

The whole line is 12

the whole line is also   x + 14  and  x + 12

  so make them equal:

      x+14     +      x +12     = 12  

       2x   + 26   = 12        now subtract 26 from both sides

               2x = - 14           now divide both sides by 2

                 x = -7

5 0
2 years ago
Read 2 more answers
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
Wich is the first step in simplifying the expression 2(x+3)+5
34kurt
Mult 2 by x you'll get 2x then 2 by 3 you'll get 6 and add 6 to 5, after that you'll get 11 it should look like this 2x+11.
7 0
4 years ago
Read 2 more answers
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