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3 years ago

Math help!

Mathematics

1 answer:

Alchen [17]3 years ago

8 0

Answer: 1675 Ft. = Distance from

fire to the windmill.

Step-by-step explanation:

tan8 = 140/X1.

X1 = 140/tan8 = 996.2 Ft.

tan3 = 140/(996+X2)

996+X2 = 140/tan3 = 2671

X2 = 2671-996 = 1675 Ft. = Distance from

fire to windmill.

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Talk to me. My dad is vincent afton my mom is vanny, my uncles are henry and william, my auntie is clara, My cousins are goldie

Shkiper50 [21]

Answer:

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Step-by-step explanation:

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2 years ago

A portion of a map with cities a and b is shown.The map uses a scale of 1 inch to 30 miles. A new map has a scale of 2 inches=15

Alina [70]

Answer:

22.5miles

Step-by-step explanation:

For Map 1;

1 inches = 30miles

For Map 2:

2 inches = 15miles

1 inches = x

2x = 1 *15

x = 15/2

x = 7.5miles

This means that 1 inch represents 7.5miles on Map 2'

The distance between the two cities in miles = 30miles - 7.5miles

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2 years ago

Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va

Burka [1]

Answer:

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

The polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

$\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)$

with

$\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}$

for some c in the interval (-x, x)

In the particular case f

f(x)=cos(x)

we have

$\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)$

therefore

$\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0$

and the polynomial approximation of T5(x) of cos(x) would be

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

$\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}$

for some c in (-x,x). So

$\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}$

and we must find the values of x for which

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652$

Working this inequality out, we find

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815$

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

8 0

2 years ago

What is the perimeter of rhombus LMNO? 20 units 24 units 40 units 48 units

NeTakaya

The little lines on each side of the rhombus mean that all the sides are the same length.

We can set line LM and MN equal to solve for X, then we can solve the length of a side.

3x-3 = x+7

Add 3 to each side:
3x = x +10

Subtract x from each side:

2x = 10

Divide both sides by 2:
x = 10/2
x = 5

Now we have the value for x, replace x in one of the side formulas:

x +7 = 5+7 = 12

Each side = 12 units.

The perimeter would be 12 + 12 + 12 + 12 = 48 units.

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2 years ago

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Where are the measurements?

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3 years ago

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