Answer:
m∠C=28°, m∠A=62°, AC=34.1 units
Step-by-step explanation:
Given In ΔABC, m∠B = 90°, , and AB = 16 units. we have to find m∠A, m∠C, and AC.
As, cos(C)={15}/{17}
⇒ angle C=cos^{-1}(\frac{15}{17})=28.07^{\circ}\sim28^{\circ}
By angle sum property of triangle,
m∠A+m∠B+m∠C=180°
⇒ m∠A+90°+28°=180°
⇒ m∠A=62°
Now, we have to find the length of AC
sin 28^{\circ}=\frac{AB}{AC}
⇒ AC=\frac{16}{sin 28^{\circ}}=34.1units
The length of AC is 34.1 units
Answer:
Part 1: 
or 
Part 2: 
Step-by-step explanation:
To start things off, you must put everything on the opposite side of y so that you only have the y variable left.
In this case, put x onto the other side, making it negative since it's flipped, so you get 
(since the x is negative you're using subtraction from 11).
Next, divide both sides by 5 to make 5y into y:
Then in this case you turn y=f(x), but the system did it for you, so all you have to put is .
For the second part you must replace x with 2 since you need to find f(2).
Your answer for part 2 is .
Volume is legnth times widht times height
lenght=2x-1
width=x-2
height=x+1
multiply all together
use mass distributive property
distributive=a(b+c)=ab+ac so extending that
(a+c)(c+c)=(a+b)(c)+(a+b)(d) then keep distributing so
(2x-1)(x-2)(x+1)
do each one seperately
do the first two first and put the other one (x+1) to the side for later
(2x-1)(x-2)=(2x-1)(x)+(2x-1)(-2)=(2x^2-x)+(-4x+2)=2x^2-5x+2
then do the other one
(x+1)(2x^2-5x+2)=(x)(2x^2-5x+2)+(1)(2x^2-5x+2)=(2x^3-5x^2+2x)+(2x^2-5x+2)=2x^3-3x^2-3x+2
the lasst form is 2x^3-3x^2-3x+2
