Answer:
the answer is A
Step-by-step explanation:
10C6 = 10! / (6! * 4!) = 10 * 9 * 8 * 7 / 4! = 5040/24 = 210
Answer:
It’s already in expanded form.
Step-by-step explanation:
The original number is 54,025.
Expanded form is when you kind of “take apart” a number. You basically break it apart.
Example: Write 24,382 in expanded form.
Correct way: 20,000 + 4,000 + 300 + 80 + 2
Incorrect way: 2 + 4 + 3 + 8 + 2 (that would equal 19, which is way off from 24,382).
I hope this helps, let me know if you have any questions!
The remainder theorem says that dividing a polynomial <em>f(x)</em> by a 1st-degree polynomial <em>g(x)</em> = <em>x</em> - <em>c</em> leaves a remainder of exactly <em>f(c)</em>.
(a) With <em>f(x)</em> = <em>px</em> ³ + 4<em>x</em> - 10 and <em>d(x)</em> = <em>x</em> + 3, we have a remainder of 5, so
<em>f</em> (-3) = <em>p</em> (-3)³ + 4(-3) - 10 = 5
Solve for <em>p</em> :
-27<em>p</em> - 12 - 10 = 5
-27<em>p</em> = 27
<em>p</em> = -1
(b) With <em>f(x)</em> = <em>x</em> + 3<em>x</em> ² - <em>px</em> + 4 and <em>d(x)</em> = <em>x</em> - 2, we have remainder 8, so
<em>f</em> (2) = 2 + 3(2)² - 2<em>p</em> + 4 = 8
-2<em>p</em> = -10
<em>p</em> = 5
(you should make sure that <em>f(x)</em> was written correctly, it's a bit odd that there are two <em>x</em> terms)
(c) <em>f(x)</em> = 2<em>x</em> ³ - 4<em>x</em> ² + 6<em>x</em> - <em>p</em>, <em>d(x)</em> = <em>x</em> - 2, <em>R</em> = <em>f</em> (2) = 18
<em>f</em> (2) = 2(2)³ - 4(2)² + 6(2) - <em>p</em> = 18
12 - <em>p</em> = 18
<em>p</em> = -6
The others are done in the same fashion. You would find
(d) <em>p</em> = 14
(e) <em>p</em> = -4359
(f) <em>p</em> = 10
(g) <em>p</em> = -13/2 … … assuming you meant <em>f(x)</em> = <em>x</em> ⁴ + <em>x</em> ³ + <em>px</em> ² + <em>x</em> + 20
Find the equation of a line given that you know points it passes through
<em>infinitely many solutions</em>
- <em>Step-by-step explanation:</em>
<em>Hi there !</em>
<em>3(y - 2) = 3y - 6</em>
<em>3y - 6 = 3y - 6 | + 6</em>
<em>3y = 3y </em>
<em>true ∀ y ∈ R</em>
<em>Good luck !</em>
