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3 years ago

Helpppppp solve it find the area please

Mathematics

1 answer:

natulia [17]3 years ago

5 0

One way is to find the area of the shape as if it was a whole square instead of an H shape.

That is (125+75+125) x (345), which is 112125 ft^2.

From this, let's subtract the areas of the cut-outs.

The cutouts are squares and both the same size. One side of it is 112 feet, so the area of one is 112x112=12544. So, the two squares have an area of 25088.

Now subtract that from the original total. 112125-25088=87037. <<that is the area.

You might want to double check my work.

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Step-by-step explanation:

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3 years ago

Help!!!!! Please!!!!!!!! Math!!!!!! Quickly!!!!!! Brainiest answer to correct answer!!!!!!!!

muminat

The answer should be 103. Because the interior angles of all triangles add up to 180 degrees, you can just add together the two known interior angles; subtract them from 180; and then find the supplement.

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3 years ago

Read 2 more answers

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o-na [289]

Answer:

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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

The incline of a roller coaster is 2 times as long as its elevation, and the horizontal length of the roller coaster is 11 m mor

Gelneren [198K]

Let
x-------> the horizontal length of the roller coaster
y------> the elevation of the roller coaster
c------> the incline of a roller coaster

we know that
applying the Pythagoras theorem
c²=x²+y²-----> equation 1
c=2y-----> equation 2
x=11+y----> equation 3

substitute equation 2 and equation 3 in equation 1
[2y]²=(11+y)²+y²
4y²=11²+22y+y²+y²
2y²-22y-121=0

using a graph tool-----> to resolve the second order equation
see the attached figure

the solution is
y=15.026 m

c=2*y-----> c=2*15.026----> c=30.052 m

the answer is
30.05 m

7 0

3 years ago