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2 years ago

Ok so- {[Math from BIM]} ss below

Mathematics

1 answer:

Mice21 [21]2 years ago

6 0

Answer:

2 real solutions

Step-by-step explanation:

x=6i, -6i

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Expand the given power by using Pascal’s triangle. (9a - 10b)^6

xxMikexx [17]

Answer:

$531441a^6-3542940a^5b+9841500a^4b^2-14580000a^3b^3+12150000a^2b^4-5400000ab^5+1000000b^6$

Step-by-step explanation:

1 n=0

1 1 n=1

1 2 1 n=2

1 3 3 1 n=3

1 4 6 4 1 n=4

1 5 10 10 5 1 n=5

1 6 15 20 15 6 1 n=6

This is where n is the exponent in

$(x+y)^n$ .

$(x+y)^6=1x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+1y^6$

Now we want to expand:

$(9a-10b)^6$ or we we can rewrite as $(9a+(-10b))^6$ .

Let's replace $x$ with $(9a)$ and $y$ with $(-10b)$ in the expansion:

$(x+y)^6=1x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+1y^6$

$((9a)+(-10b))^6$

$=1(9a)^6+6(9a)^5(-10b)+15(9a)^4(-10b)^2+20(9a)^3(-10b)^3+15(9a)^2(-10b)^4+6(9a)(-10b)^5+1(-10b)^6$

Let's simplify a bit:

$=9^6a^6-60(9)^5a^5b+15(-10)^2(9)^4a^4b^2+20(9)^3(-10)^3a^3b^3+15(9)^2(-10)^4a^2b^4+6(9)(-10)^5ab^5+(-10)^6b^6$

$=531441a^6-3542940a^5b+9841500a^4b^2-14580000a^3b^3+12150000a^2b^4-5400000ab^5+1000000b^6$

8 0

3 years ago

ram bought a second hand cycle for RS.1250 and spent rs.200 on its repair. when he sold if after few days he got RS.1250. what w

Anna [14]

Answer:

hope it helps you to please mark me brainliest

5 0

2 years ago

Explain Pe^(rt) How do I use this formula ?

Montano1993 [528]

Answer:

The equation for "continual" growth (or decay) is A = Pert, where "A", is the ending amount, "P" is the beginning amount (principal, in the case of money), "r" is the growth or decay rate (expressed as a decimal), and "t" is the time (in whatever unit was used on the growth/decay rate).

Step-by-step explanation:

<h2> Don't sweat here is a video link too </h2>

Compounding Continuously Pert Formula

https://youtu.be/dFsBfi9W7sQ

8 0

2 years ago

The function f(x) = x2 − 12x + 5 written in vertex form is f(x) = (x − 6)2 − 31. What are the coordinates of the vertex?

Artist 52 [7]

$\bf ~~~~~~\textit{parabola vertex form}
\\\\
\begin{array}{llll}
\boxed{y=a(x- h)^2+ k}\\\\
x=a(y- k)^2+ h
\end{array}
\qquad\qquad
vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\
-------------------------------\\\\
f(x)=1(x-\stackrel{h}{6})^2\stackrel{k}{-31}\qquad \qquad vertex~(6,-31)$

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3 years ago

Read 2 more answers

Can someone help me with this please

Arisa [49]

I’ve attached my work...
Hope it helps!
Math-way app for problems like number 2
Have a nice day:)

6 0

2 years ago