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3 years ago

Evaluate the following logarithms log12144 = log151 = log 0.00001 =

Mathematics

2 answers:

gladu [14]3 years ago

4 0

Answer:

1- 2

2- 0

3- -4

4- -5

Step-by-step explanation:

Mnenie [13.5K]3 years ago

3 0

Answer:

Log12144 = 2

Log151 = 0

log3(1/81) = -4

log0.00001 = -5

Step-by-step explanation:

I just done that

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What type of transformation takes the graph of f(x)=|x| to the graph of g(x)=−3+|x|?

crimeas [40]

Answer:

vertical translation of 3 units down

Step-by-step explanation:

we have

$f\left(x\right)=\left|x\right|$ ----> the parent function

The vertex of f(x) is the point (0,0)

$g\left(x\right)=-3+\left|x\right|$ ----> the transformed function

The vertex of g(x) is the point (0,-3)

The rule of the translation is

f(x) -----> g(x)

(0,0) ----> (0,-3)

(x,y) ----> (x,y-3)

That means ---> The translation is 3 units down

see the attached figure to better understand the problem

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3 years ago

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Alex777 [14]

Number 1 is 2.5, hope this helps a bit!

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3 years ago

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Harvey can run 3 1/6 km in 1/4 hour. Harvey runs at a constant rate

lyudmila [28]

Answer:

12 2/3 km/hour or 12.666 km/hour

Step-by-step explanation:

Because 3 1/6 is the rate for every quarter hour, simply multiply that rate by 4 in order to get the average speed over the course of an hour.

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3 years ago

If θ is an angle in standard position that passes through (-3, 4), find sin2θ.

julia-pushkina [17]

Answer:

Correct answer: sin 2Θ = - 24/25

Step-by-step explanation:

If under the standard position you think that the first arm (side) belongs to the positive direction of the x axis and the second one passes through a given point then it is:

if we form a right triangle with sides 3 and 4 then the hypotenuse is 5.

sin Θ = 4/5 and cos Θ = - 3/5

we know that the formula for double the value of the angle is:

sin 2Θ = 2 sinΘ cosΘ = 2 · 4/5 · ( - 3/5) = - 24/25

sin 2Θ = - 24/25

God is with you!!!

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3 years ago

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0

3 years ago