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3 years ago

Evaluate each logarithm. 100 Points for all questions.

Mathematics

2 answers:

mestny [16]3 years ago

6 0

Answer:

19. 1/2

21. 47

23. 3

25. 6

27. 5/6

Step-by-step explanation:

all you had to do is use a calculator

seraphim [82]3 years ago

3 0

Basic properties that will be used:

$\log_ab^n=n\log_ab$

$x^{1/n}=\sqrt[n]x$

$\log_aa=1$

All of these expression can be written to match on or more of the patterns above.

$19.\quad\log_5\sqrt[4]{25}=\log_5(5^2)^{1/4}=\dfrac14\log_55^2=\dfrac12\log_55=\dfrac12$

$21.\quad9\ln e^3+4\ln e^5=27\ln e+20\ln e=27+20=47$

$23.\quad2\log_3\sqrt{27}=2\log_3(3^3)^{1/2}=\log_33^3=3\log_33=3$

$25.\quad4\log_2\sqrt8=4\log_2(2^3)^{1/2}=2\log_22^3=6\log_22=6$

$27.\quad\log_3\sqrt[6]{243}=\log_3(3^5)^{1/6}=\dfrac16\log_33^5=\dfrac56\log_33=\dfrac56$

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astra-53 [7]

we have

$(2.3)(5.06)$

we know that

$2.3=\frac{23}{10}$

$5.06=\frac{506}{100}$

$(2.3)(5.06)=(\frac{23}{10})(\frac{506}{100})$

$=\frac{11,638}{1,000}$

$=11.638$

the answer is in thousandths, because the denominator of the multiplication of the two fractions is one thousand

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3 years ago

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Answer quickly please

KIM [24]

The correct answer is C.

You can tell this by factoring the equation to get the zeros. To start, pull out the greatest common factor.

f(x) = x^4 + x^3 - 2x^2

Since each term has at least x^2, we can factor it out.

f(x) = x^2(x^2 + x - 2)

Now we can factor the inside by looking for factors of the constant, which is 2, that add up to the coefficient of x. 2 and -1 both add up to 1 and multiply to -2. So, we place these two numbers in parenthesis with an x.

f(x) = x^2(x + 2)(x - 1)

Now we can also separate the x^2 into 2 x's.

f(x) = (x)(x)(x + 2)(x - 1)

To find the zeros, we need to set them all equal to 0

x = 0

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x = -2

x - 1 = 0

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3 years ago

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Marsha wants to determine the vertex of the quadratic function f(x) = x2 – x + 2. What is the function’s vertex?

BaLLatris [955]

Answer:

$Vertex = (\frac{1}{2},\frac{7}{4})$

Step-by-step explanation:

Given

$f(x) = x^2 - x +2$

Required

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We have:

$f(x) = x^2 - x +2$

First, we express the equation as:

$f(x) = a(x - h)^2 +k$

Where

$Vertex = (h,k)$

So, we have:

$f(x) = x^2 - x +2$

--------------------------------------------

Take the coefficient of x: -1

Divide by 2: (-1/2)

Square: (-1/2)^2

Add and subtract this to the equation

--------------------------------------------

$f(x) = x^2 - x +2$

$f(x) = x^2 - x + (-\frac{1}{2})^2+2 -(-\frac{1}{2})^2$

$f(x) = x^2 - x + \frac{1}{4}+2 -\frac{1}{4}$

Expand

$f(x) = x^2 - \frac{1}{2}x- \frac{1}{2}x + \frac{1}{4}+2 -\frac{1}{4}$

Factorize

$f(x) = x(x - \frac{1}{2})- \frac{1}{2}(x - \frac{1}{2})+2 -\frac{1}{4}$

Factor out x - 1/2

$f(x) = (x - \frac{1}{2})(x - \frac{1}{2})+2 -\frac{1}{4}$

$f(x) = (x - \frac{1}{2})^2+2 -\frac{1}{4}$

$f(x) = (x - \frac{1}{2})^2+ \frac{8 -1 }{4}$

$f(x) = (x - \frac{1}{2})^2+ \frac{7}{4}$

Compare to: $f(x) = a(x - h)^2 +k$

$h = \frac{1}{2}$

$k = \frac{7}{4}$

Hence:

$Vertex = (\frac{1}{2},\frac{7}{4})$

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dmitriy555 [2]

Answer:

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