All categories

4 years ago

What is the arc length of a circle that has an 8-inch radius and a central angle that is 95 degrees?

1 answer:

7 0

$\bf \textit{arc's length}\\\\
s=\cfrac{\theta \pi r}{180}~~
\begin{cases}
r=radius\\
\theta =angle~in\\
\qquad radius\\
------\\
r=8\\
\theta =95
\end{cases}\implies s=\cfrac{(95)(\pi )(8)}{180}\implies s=\cfrac{38\pi }{9}$

You might be interested in

At an ice cream shop, a customer can make a sundae that includes ice cream, one candy topping, and one liquid topping from the l

Elden [556K]

Answer:

Step-by-step explanation:

Regular

Candy

Liquid

Both

8 0

3 years ago

Which table has a constant of proportionality between y and x of 12?

konstantin123 [22]

Answer:

Table A

Step-by-step explanation:

If you do delta method (Change in y and x)

For example, 120-24

______ = 96/8

10-2

Then you divide both the numerator and denominator by the denominator (which is 8) and that gets us to 12 as our answer. Do the same method for all of the numbers in table A.

Ps: I hope that I was helpful, and you are welcome.

6 0

3 years ago

Solving the right triangle ‼️ (Round to the nearest tenth) can someone help me find A,a, and b❓

Nataliya [291]

A is 40 degrees imma edit this or do it in comments the other ones

3 0

4 years ago

Read 2 more answers

What's f(g(4)) if f(x) = 3x2 - 3x + 6 and g(x) = 2x?<br> OA) 210<br> B) 174<br> C) 12<br> OD) 24

BaLLatris [955]

Answer:

Step-by-step explanation:

g(4) = 8

f(8) = 3(8)^2 - 3(8) 6 = 174

8 0

3 years ago

Use Cramer’s rule to solve for x: x + 4y − z = −14 5x + 6y + 3z = 4 −2x + 7y + 2z = −17

V125BC [204]

Looks like the system is

x + 4y - z = -14

5x + 6y + 3z = 4

-2x + 7y + 2z = -17

or in matrix form,

$\mathbf{Ax} = \mathbf b \iff \begin{bmatrix} 1 & 4 & -1 \\ 5 & 6 & 3 \\ -2 & 7 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -14 \\ 4 \\ -17 \end{bmatrix}$

Cramer's rule says that

$x_i = \dfrac{\det \mathbf A_i}{\det \mathbf A}$

where $x_i$ is the solution for i-th variable, and $\mathbf A_i$ is a modified version of $\mathbf A$ with its i-th column replaced by $\mathbf b$ .

We have 4 determinants to compute. I'll show the work for det(A) using a cofactor expansion along the first row.

$\det \mathbf A = \begin{vmatrix} 1 & 4 & -1 \\ 5 & 6 & 3 \\ -2 & 7 & 2 \end{vmatrix}$

$\det \mathbf A = \begin{vmatrix} 6 & 3 \\ 7 & 2 \end{vmatrix} - 4 \begin{vmatrix} 5 & 3 \\ -2 & 2 \end{vmatrix} - \begin{vmatrix} 5 & 6 \\ -2 & 7 \end{vmatrix}$

$\det \mathbf A = ((6\times2)-(3\times7)) - 4((5\times2)-(3\times(-2)) - ((5\times7)-(6\times(-2)))$

$\det\mathbf A = 12 - 21 - 40 - 24 - 35 - 12 = -120$

The modified matrices and their determinants are

$\mathbf A_1 = \begin{bmatrix} -14 & 4 & -1 \\ 4 & 6 & 3 \\ -17 & 7 & 2\end{bmatrix} \implies \det\mathbf A_1 = -240$

$\mathbf A_2 = \begin{bmatrix} 1 & -14 & -1 \\ 5 & 4 & 3 \\ -2 & -17 & 2 \end{bmatrix} \implies \det\mathbf A_2 = 360$

$\mathbf A_3 = \begin{bmatrix} 1 & 4 & -14 \\ 5 & 6 & 4 \\ -2 & 7 & -17 \end{bmatrix} \implies \det\mathbf A_3 = -480$

Then by Cramer's rule, the solution to the system is

$x = \dfrac{-240}{-120} \implies \boxed{x = 2}$

$y = \dfrac{360}{-120} \implies \boxed{y = -3}$

$z = \dfrac{-480}{-120} \implies \boxed{z = 4}$

5 0

2 years ago