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3 years ago

Find the measure of 2. 63° 117° 62 = [?] Enter

Mathematics

1 answer:

forsale [732]3 years ago

7 0

Answer:

63°

Step-by-step explanation:

Angle <2 and the angle with measure of 63 are alternate exterior angles and has same measurement.

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The quotient of 2 and t

tino4ka555 [31]

Answer: 2t/12

Step-by-step explanation:The quotient of is the result of dividing two terms

The first term is - Twice a number

4 0

3 years ago

From the hay loft door, Ted sees his dog on the ground. The angle of depression of the dog is 40º. Ted's eye level is 16 feet ab

Vlad1618 [11]

Answer: 19 feet

Step-by-step explanation:

Hi, since the situation forms a right triangle (see attachment) we have to apply the next trigonometric function.

Tan α = opposite side / adjacent side

Where α is the angle of depression of the dog, the opposite side (16) is Ted's eye level above the ground, and the adjacent side (x) is the distance between the dog and the barn door.

Replacing with the values given:

tan40 = 16/x

Solving for x

x =16/tan40

x= 19 ft

Feel free to ask for more if needed or if you did not understand something.

5 0

3 years ago

What is the slope of x+y=7,x-y=8

slava [35]

Step-by-step explanation:

The slop is M= 1÷x-y-xy

5 0

3 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

3 years ago

A distribution x is known to have a mean value of 5 and a standard deviation of 5. what is its mean square value (i.e., the expe

telo118 [61]

$\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$
$\implies 5^2=\mathbb E(X^2)-5^2$
$\implies\mathbb E(X^2)=50$

8 0

3 years ago