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2 years ago

Answer quickly only if you know

Mathematics

1 answer:

Neko [114]2 years ago

7 0

The figure has 5 sides, so it must be a pentagon. Pentagon’s interior angles sum up to 540°.

Therefore:

132+90+3x=540

222+3x=540

3x=318

x=106

Checking work, where x=106

222+3(106)=540.

So x°=106°

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Please help me!

Alexxandr [17]

A right triangle has one leg with unknown length, the other leg with length of 5 m, and the hypotenuse with length 13 times sqrt 5 m.

We can use the Pythagorean formula to find the other leg of the right triangle.

a²+b²=c²

Where a and b are the legs of the triangle and c is the hypotenuse.

According to the given problem,

one leg: a= 5m and hypotenuse: c=13√5 m.

So, we can plug in these values in the above equation to get the value of unknown side:b. Hence,

5²+b²=(13√5)²

25 + b² = 13²*(√5)²

25 + b² = 169* 5

25+ b² = 845

25 + b² - 25 = 845 - 25

b² = 820

b =√ 820

b = √(4*205)

b = √4 *√205

b = 2√205

b= 2* 14.32

b = 28.64

So, b= 28.6 (Rounded to one decimal place)

Hence, the exact length of the unknown leg is 2√205m or 28.6 m (approximately).

5 0

3 years ago

A parabola can be drawn given a focus of (-11, -2) and a directrix of x= -3

fgiga [73]

Check the picture below, so the parabola looks more or less like that.

now, the vertex is half-way between the focus point and the directrix, so that puts it where you see it in the picture, and the horizontal parabola is opening to the left-hand-side, meaning that the distance "P" is negative.

$\textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\supset}\qquad \stackrel{"p"~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\begin{cases} h=-7\\ k=-2\\ p=-4 \end{cases}\implies 4(-4)[x-(-7)]~~ = ~~[y-(-2)]^2 \\\\\\ -16(x+7)=(y+2)^2\implies x+7=-\cfrac{(y+2)^2}{16}\implies x=-\cfrac{1}{16}(y+2)^2-7$

8 0

2 years ago

Find a power series for the function, centered at c, and determine the interval of convergence. f(x) = 9 3x + 2 , c = 6

san4es73 [151]

Answer:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ........$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

Step-by-step explanation:

Given

$f(x)= \frac{9}{3x+ 2}$

$c = 6$

The geometric series centered at c is of the form:

$\frac{a}{1 - (r - c)} = \sum\limits^{\infty}_{n=0}a(r - c)^n, |r - c| < 1.$

Where:

$a \to$ first term

$r - c \to$ common ratio

We have to write

$f(x)= \frac{9}{3x+ 2}$

In the following form:

$\frac{a}{1 - r}$

So, we have:

$f(x)= \frac{9}{3x+ 2}$

Rewrite as:

$f(x) = \frac{9}{3x - 18 + 18 +2}$

$f(x) = \frac{9}{3x - 18 + 20}$

Factorize

$f(x) = \frac{1}{\frac{1}{9}(3x + 2)}$

Open bracket

$f(x) = \frac{1}{\frac{1}{3}x + \frac{2}{9}}$

Rewrite as:

$f(x) = \frac{1}{1- 1 + \frac{1}{3}x + \frac{2}{9}}$

Collect like terms

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2}{9}- 1}$

Take LCM

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2-9}{9}}$

$f(x) = \frac{1}{1 + \frac{1}{3}x - \frac{7}{9}}$

So, we have:

$f(x) = \frac{1}{1 -(- \frac{1}{3}x + \frac{7}{9})}$

By comparison with: $\frac{a}{1 - r}$

$a = 1$

$r = -\frac{1}{3}x + \frac{7}{9}$

$r = -\frac{1}{3}(x - \frac{7}{3})$

At c = 6, we have:

$r = -\frac{1}{3}(x - \frac{7}{3}+6-6)$

Take LCM

$r = -\frac{1}{3}(x + \frac{-7+18}{3}+6-6)$

r = -\frac{1}{3}(x + \frac{11}{3}+6-6)

So, the power series becomes:

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}ar^n$

Substitute 1 for a

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}1*r^n$

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}r^n$

Substitute the expression for r

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}(-\frac{1}{3}(x - \frac{7}{3}))^n$

Expand

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}[(-\frac{1}{3})^n* (x - \frac{7}{3})^n]$

Further expand:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ................$

The power series converges when:

$\frac{1}{3}|x - \frac{7}{3}| < 1$

Multiply both sides by 3

$|x - \frac{7}{3}|$

Expand the absolute inequality

$-3 < x - \frac{7}{3}$

Solve for x

$\frac{7}{3} -3 < x$

Take LCM

$\frac{7-9}{3} < x$

$-\frac{2}{3} < x$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

6 0

2 years ago

Determine whether the dimensions of each rectangle described of Taylor's first rectangle.

Svetllana [295]

Step-by-step explanation:

yes

that is the correct answer

8 0

1 year ago

Uppose it costs $10 to roll a pair of dice. you get paid the sum of the numbers in dollars that appear on the dice. what is the

pentagon [3]

At most $12, at least $2 a probability chart is best to solve for this

6 0

3 years ago

Answer quickly only if you know ​

Answer quickly only if you know