$x=\sqrt{880-8x}\\
D:880-8x\geq0 \wedge x\geq0\\
D:8x\leq880 \wedge x\geq0\\
D:x\leq 110 \wedge x\geq0\\
D:x\in\langle0,110\rangle\\
x^2=880-8x\\
x^2+8x-880=0\\
x^2+8x+16-896=0\\
(x+4)^2=896\\
x+4=8\sqrt{14} \vee x+4=-8\sqrt{14}\\
x=-4+8\sqrt{14} \vee x=-4-8\sqrt{14}\\
\\ x=-4-8\sqrt{14}\not \in D \Rightarrow \boxed{x=-4+8\sqrt{14}}$

5 0

2 years ago

Evaluate 11c3 and 11p4

In-s [12.5K]

$_{n}C_{r} = \frac{n!}{r!(n-r)!} \\ \\ _{11}C_{3}= \frac{11!}{3!*(11-3)!} =\frac{11*10*9*8*7*6*5*4*3*2*1}{3*2*1*8*7*6*5*4*3*2*1} =\frac{11*10*9}{3*2} =165 \\ \\_{11}C_{4}=\frac{11!}{4!(11-4)!} =\frac{11*10*9*8*7!}{4!*7!} =\frac{11*10*9*8}{4*3*2*1} =330$

7 0

3 years ago

Read 2 more answers

A bag contains 3 red marbles and 4 blue marbles. Two marbles are drawn at random without replacement. what is the probability th

Yuliya22 [10]

Probability (first marble red and second marble blue) = (3/7) × (4/6) = 2/7

7 0

2 years ago