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Mrac [35]
3 years ago
8

What will the coordinate K(3, - 3) be after the translation 8 units to the left and 7 units up.

Mathematics
2 answers:
natita [175]3 years ago
8 0

Answer:

(-5, 4)

Step-by-step explanation:

To find the new coordinate, subtract 8 from the x value and add 7 to the y value:

3 - 8 = -5

-5 will be the new x coordinate

-3 + 7 = 4

4 will be the new y coordinate

So, the coordinate K will be (-5, 4) after the translation

erastovalidia [21]3 years ago
4 0

Answer:

(-5,4)

Step-by-step explanation:

3-8=-5

-3+7=4

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Alina [70]

Answer:

The area of the sidewalk is 144.44 m².

The 2-m wide walk adds 4 m to the diameter, making it 21+4=25.

Since the radius is half the diameter, r = 25/2 = 12.5.

The area of the entire bed with walkway is 3.14(12.5)² = 490.625 m².

The diameter of the bed is 21, so the radius is 21/2 = 10.5.

The area of just the flower bed is 3.14(10.5)² = 346.185.

The difference between the two is 490.625-346.185 = 144.44 m².  This is the area of the walkway.

<em>Have a Great Day!</em>

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2 years ago
What are the coordinates of the vertex of the parabola represent by the equation y=-5x^2+30x-25
Jobisdone [24]

Answer:

The coordinate of the  vertex of the parabola is (h,k) =  (3,20)

Step-by-step explanation:

The given equation of the parabola is y=-5x^2+30x-25

Now, the General Parabolic Equation is of the form:

y = a (x-h)^2 + k, then the vertex is the point (h, k).

Now, to covert the given equation in the standard form:

y=-5x^2+30x-25   = -5(x^2  -6x   + 5)

Using the COMPLETE THE SQUARE METHOD,

-5(x^2  -6x   + 5)  = -5(x^2  -6x  +  (3)^2 +  5  - (3)^2)\\\implies -5(((x^2  -6x +(3)^2)  + 5 - 9)\\= -5((x-3)^3 -4)\\= -5(x-3)^2 + 20

ory = -5(x-3)^2 + 20

⇒The general formed equation of the given parabola is

y  = -5(x-3)^2 + 20

Comparing this with general form, we get

h = 3, k = 20

Hence, the coordinate of the  vertex of the parabola is (h,k) =  (3,20)

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Answer:

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Step-by-step explanation:

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Read 2 more answers
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
2 years ago
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