$f(x)=x\sqrt{16-x^2}+c;\ P(3;\ 6)\to x=3\ and\ y=6\\\\subtitute\\\\6=3\sqrt{16-3^2}+c\\\\3\sqrt{16-9}+c=6\\\\3\sqrt7+c=6\ \ \ \ |subtract\ 3\sqrt7\ from\ both\ sides\\\\\boxed{c=6-3\sqrt7}$

4 0

4 years ago

Find the area of a square with side lengths of 40x^2y^3

Korolek [52]

Answer:

grjik

Step-by-step explanation:

6 0

4 years ago

tammy typed 40 pages of a report on Monday and 3/8 of the report on Tuesday. She completed the remaining 1/3 of the report on We

sergeinik [125]

Answer:

Step-by-step explanation:

R = report length

R = 40 + (3/8)R + (1/3)R

3/8 = 3*3/8*3 = 9/24

1/3 = 8*1 / 3*8 = 8/34

R = 40 + (9/24)R + (8/24)R

R = 40 + (17/24)R Subtract 17/24 from the left

R-17/24 R = 40

7/24 R = 40 Multiply both sides by 24

7R = 40 * 24

7R = 960 Divide by 7

R = 960/7

R = 137.1 which breaks the rounding rule and becomes 138 because something has to be typed on page 138

6 0

4 years ago

HELP PLEASE ILL GIVE THE BRANLIEST THING IF U DO, PLEASEE

bezimeni [28]

Answer:

The y-intercept is 5, and the slope is -3/2, y=5

Step-by-step explanation:

8 0

2 years ago