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aleksley [76]
3 years ago
6

How do I evaluate log 1/32?

Mathematics
2 answers:
stira [4]3 years ago
4 0
On a graphing calculator, hit Math, Alpha, Math. Then input your logarithmic function. Hope this helped! Good luck :)
Mandarinka [93]3 years ago
4 0
Answer:

1/5

Explanation:

Use <span><span><span>logb</span>a</span>=<span><span><span>logc</span>a</span><span><span>logc</span>b</span></span></span>

Here,

<span><span>log<span>12</span></span><span>(<span>132</span>)</span></span>

<span>=<span><span>log<span>(<span>132</span>)</span></span><span>/log<span>(<span>12</span>)</span></span></span></span>

<span>=<span><span>−<span>log2</span></span><span>−<span>log32</span></span></span></span>

<span>=<span><span>log2</span><span><span> /log2</span>5</span></span></span>

<span>=<span><span>log 2</span><span>5 /<span>log2</span></span></span></span>

<span>=<span>1<span>/5</span></span></span>
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58.8\ mm^2

Step-by-step explanation:

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Let v be a vector with initial point (-5, -11) and terminal point (6,2). What is v?
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Solve with cramer's rule x+2y+3z=11, 2x+y+2z=10, 3x+2y+z=9
nalin [4]

Answer:

x = 2 , y = 0 , z = 3

Step-by-step explanation:

Cramer's rule is a rule through which we can find the solution of linear equation.

we have the three linear equations as

x+2y+3z=11

2x+y+2z=10

3x+2y+z=9

AX=B  

A: coefficient matrix

X= unknown vectors(x,y,z)

D = values of the linear equation (11 , 10 , 9)

now we find the determinant of the given linear equation

determinant of the matrix will be

A = \left[\begin{array}{ccc}1&2&3\\2&1&2\\3&2&1\end{array}\right]  = 1(1-4) - 2(2-6) + 3(4 - 3)

                    = 1(-3) - 2(-4) + 3(1)

                    = -3+8+3 = 8

also D\neq 0

so the determinant is Non zero we can apply Cramer's rule

we will be replacing the first column of the coefficient matrix A with the values of D

by replacing the first column we will get the value of the variable 'x'

Dx =  \left[\begin{array}{ccc}11&2&3\\10&1&2\\9&2&1\end{array}\right]   = 11(1-4) -2(10-18) + 3(20-9) = -33+16+33 = 16

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so we have the solution as

x = 2 , y = 0 , z = 3

Therefore the solution for the given linear equations is (2,0,3).

3 0
3 years ago
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