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Alex777 [14]
3 years ago
14

Evaluate the logarithmic expression log6 1

Mathematics
1 answer:
AlekseyPX3 years ago
3 0

Answer:

Logarithm base  6  of  1  is<u>  0</u>.

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PLSSSS HELPPPPPPPPPPPPPPPPPP
MArishka [77]

Answer:

\huge\boxed{\sf Option \ 1}

Step-by-step explanation:

People surveyed = 42

People who approved = 25

<u>So, the sample is:</u>

25 residents who approve.

<u>And population:</u>

42 residents surveyed.

\rule[225]{225}{2}

3 0
2 years ago
Hahaha help:)))))))))
Viefleur [7K]

Answer:

1332 i think is how it will be writen

Step-by-step explanation:

8 0
3 years ago
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What is the equation in slope-intercept form of the line that passes through the points (2,-3)
Fittoniya [83]

Answer:

y=1x-6

Step-by-step explanation:

luckily i love math, plz put as brainliest

6 0
2 years ago
−2x=x^2−6
Iteru [2.4K]

Step-by-step explanation:

Example 1

Solve the equation x3 − 3x2 – 2x + 4 = 0

We put the numbers that are factors of 4 into the equation to see if any of them are correct.

f(1) = 13 − 3×12 – 2×1 + 4 = 0 1 is a solution

f(−1) = (−1)3 − 3×(−1)2 – 2×(−1) + 4 = 2

f(2) = 23 − 3×22 – 2×2 + 4 = −4

f(−2) = (−2)3 − 3×(−2)2 – 2×(−2) + 4 = −12

f(4) = 43 − 3×42 – 2×4 + 4 = 12

f(−4) = (−4)3 − 3×(−4)2 – 2×(−4) + 4 = −100

The only integer solution is x = 1. When we have found one solution we don’t really need to test any other numbers because we can now solve the equation by dividing by (x − 1) and trying to solve the quadratic we get from the division.

Now we can factorise our expression as follows:

x3 − 3x2 – 2x + 4 = (x − 1)(x2 − 2x − 4) = 0

It now remains for us to solve the quadratic equation.

x2 − 2x − 4 = 0

We use the formula for quadratics with a = 1, b = −2 and c = −4.

We have now found all three solutions of the equation x3 − 3x2 – 2x + 4 = 0. They are: eftirfarandi:

x = 1

x = 1 + Ö5

x = 1 − Ö5

Example 2

We can easily use the same method to solve a fourth degree equation or equations of a still higher degree. Solve the equation f(x) = x4 − x3 − 5x2 + 3x + 2 = 0.

First we find the integer factors of the constant term, 2. The integer factors of 2 are ±1 and ±2.

f(1) = 14 − 13 − 5×12 + 3×1 + 2 = 0 1 is a solution

f(−1) = (−1)4 − (−1)3 − 5×(−1)2 + 3×(−1) + 2 = −4

f(2) = 24 − 23 − 5×22 + 3×2 + 2 = −4

f(−2) = (−2)4 − (−2)3 − 5×(−2)2 + 3×(−2) + 2 = 0 we have found a second solution.

The two solutions we have found 1 and −2 mean that we can divide by x − 1 and x + 2 and there will be no remainder. We’ll do this in two steps.

First divide by x + 2

Now divide the resulting cubic factor by x − 1.

We have now factorised

f(x) = x4 − x3 − 5x2 + 3x + 2 into

f(x) = (x + 2)(x − 1)(x2 − 2x − 1) and it only remains to solve the quadratic equation

x2 − 2x − 1 = 0. We use the formula with a = 1, b = −2 and c = −1.

Now we have found a total of four solutions. They are:

x = 1

x = −2

x = 1 +

x = 1 −

Sometimes we can solve a third degree equation by bracketing the terms two by two and finding a factor that they have in common.

6 0
3 years ago
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6th grade math ! help me please :)
malfutka [58]
1. N + 2; n = 6: the answer is 8 because n=6 so we sub n with that... and we get 6+2 and that’s 8.

2.5f; where f=4: the answer is 20 because when a variable is directly next to a number it is multiplied by that number so we will replace f with 4 and our equation is now 5(4) or 5•4 and both are equivalent to 20.

3. 7b-2; where b=5: the answer is 33 because 7 multiplied by 5(b) minus 2= 7(5)-2 or 7•5-2= 33 because you will multiply 7 by 5 and get 33 then you will subtract by 2 and get 33.

Hope this helps!!!
7 0
3 years ago
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