Answer:
The new account balance is $40000
Step-by-step explanation:
<em>The number line is not given. However, the question can be solved without it.</em>
Given
Required
Determine the new balance
Hence:
<em>The new account balance is $40000</em>
The decimal approximation for the trigonometric function sin 28°48' is
Given the trigonometric function is sin 28°48'
The ratio between the adjacent side and the hypotenuse is called cos(θ), whereas the ratio between the opposite side and the hypotenuse is called sin(θ). The sin(θ) and cos(θ) values for a given triangle are constant regardless of the triangle's size.
To solve this, we are going to convert 28°48' into degrees first, using the conversion factor 1' = 1/60°
sin (28°48') = sin(28° ₊ (48 × 1/60)°)
= sin(28° ₊ (48 /60)°)
= sin(28° ₊ 4°/5)
= sin(28° ₊ 0.8°)
= sin(28.8°)
= 0.481753
Therefore sin (28°48') is 0.481753.
Learn more about Trigonometric functions here:
brainly.com/question/25618616
#SPJ9
Arabia because she has the largest numbers
Check the picture below.
![~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{-9})\qquad B(\stackrel{x_2}{8}~,~\stackrel{y_2}{0}) ~\hfill AB=\sqrt{[ 8- 1]^2 + [ 0- (-9)]^2} \\\\\\ AB=\sqrt{7^2+(0+9)^2}\implies AB=\sqrt{7^2+9^2}\implies \boxed{AB=\sqrt{130}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~%5Chfill%20%5Cstackrel%7B%5Ctextit%7B%5Clarge%20distance%20between%202%20points%7D%7D%7Bd%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%7D~%5Chfill~%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20A%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B-9%7D%29%5Cqquad%20B%28%5Cstackrel%7Bx_2%7D%7B8%7D~%2C~%5Cstackrel%7By_2%7D%7B0%7D%29%20~%5Chfill%20AB%3D%5Csqrt%7B%5B%208-%201%5D%5E2%20%2B%20%5B%200-%20%28-9%29%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20AB%3D%5Csqrt%7B7%5E2%2B%280%2B9%29%5E2%7D%5Cimplies%20AB%3D%5Csqrt%7B7%5E2%2B9%5E2%7D%5Cimplies%20%5Cboxed%7BAB%3D%5Csqrt%7B130%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![B(\stackrel{x_1}{8}~,~\stackrel{y_1}{0})\qquad C(\stackrel{x_2}{9}~,~\stackrel{y_2}{-8}) ~\hfill BC=\sqrt{[ 9- 8]^2 + [ -8- 0]^2} \\\\\\ BC=\sqrt{1^2+(-8)^2}\implies \boxed{BC=\sqrt{65}}](https://tex.z-dn.net/?f=B%28%5Cstackrel%7Bx_1%7D%7B8%7D~%2C~%5Cstackrel%7By_1%7D%7B0%7D%29%5Cqquad%20C%28%5Cstackrel%7Bx_2%7D%7B9%7D~%2C~%5Cstackrel%7By_2%7D%7B-8%7D%29%20~%5Chfill%20BC%3D%5Csqrt%7B%5B%209-%208%5D%5E2%20%2B%20%5B%20-8-%200%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20BC%3D%5Csqrt%7B1%5E2%2B%28-8%29%5E2%7D%5Cimplies%20%5Cboxed%7BBC%3D%5Csqrt%7B65%7D%7D)
now, we could check for the CA distance, however, we already know that AB ≠ BC, so there's no need.
