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2 years ago

(a+b-c)(ab-bc+ca)expand

Mathematics

1 answer:

Yuri [45]2 years ago

4 0

Answer:

a2b+ca2+ab2-b2c+bc2-c2a-abc

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A box of candy hearts contains 52 hearts, of which 19 are white, 10 are tan, 7 are pink, 3 are purple, 5 are yellow, 2 are orang

laiz [17]

<u>Answer-</u>

a. Probability that three of the candies are white = 0.29

b. Probability that three are white, 2 are tan, 1 is pink, 1 is yellow, and 2 are green = 0.006

<u>Solution-</u>

There are 19 white candies, out off which we have to choose 3.

The number of ways we can do the same process =

$\binom{19}{3} = \frac{19!}{3!16!} = 969$

As we have to draw total of 9 candies, after 3 white candies we left with 9-3 = 6, candies. And those 6 candies have to be selected from 52-19 = 33 candies, (as we are drawing candies other than white, so it is subtracted)

And this process can be done in,

$\binom{33}{6} = \frac{33!}{6!27!} =1107568$

So total number of selection = (969)×(1107568) = 1073233392

Drawing 9 candies out of 52 candies,

$\binom{52}{9} = \frac{52!}{9!43!} = 3679075400$

∴P(3 white candies) = $\frac{1073233392}{3679075400} =0.29$

Total number of ways of selecting 3 whites, 2 are tans, 1 is pink, 1 is yellow, and 2 are greens is,

$\binom{19}{3} \binom{10}{2} \binom{7}{1} \binom{5}{1} \binom{6}{2}$

$=(\frac{19!}{3!16!}) (\frac{10!}{2!8!}) (\frac{7!}{1!6}) (\frac{5!}{1!4!}) (\frac{6!}{2!4!})$

$=(969)(45)(7)(5)(15)=22892625$

Total number of selection = 3 whites + 2 are tans + 1 is pink + 1 is yellow + 2 greens = 9 candies out of 52 candies is,

$\binom{52}{9}=\frac{52!}{9!43!} =3679075400$

∴ P( 3 whites, 2 are tans, 1 is pink, 1 is yellow, 2 greens) =

$\frac{22892625}{3679075400} = 0.006$

6 0

4 years ago

What are the solutions to 7x^2+10x-3=x^2+3x

marysya [2.9K]

Answer:

x = 1 / 3

x = -1.5

Step-by-step explanation:

First move everything onto one side.

6x^2 + 7x - 3 = 0

Divide by 6:

x^2 + 7/6x - 0.5 = 0

Now, you can move the 0.5 to the other side because it is determined that this is not factorable.

x^2 + 7/6x = 0.5

Now you have to complete the square.

Take 1 / 2 of 7 / 6 and square it.

49 / 144

x^2 + 7/6x + 49/144 = 0.5 + 49/144

(x + 7 / 12)^2 = 121 / 144

Square root:

x + 7 / 12 = ±11 / 12

x = 1 / 3, x = -1.5

7 0

4 years ago

Please somebody answer this. What is the length and width of the rectangle?

mafiozo [28]

Answer:

The length of the rectangle is $11x^{4}$ and width is $(5x^{2} + 1)$ .

Step-by-step explanation:

The area of the green zone is $55x^{6}$ and the area of the purple zone is $22x^{4}$ .

So, the total shaded zone is a rectangle having area = $55x^{6} + 22x^{4}$ .

Now, factorizing the area expression we get, area = $11x^{4}(5x^{2} + 2)$ .

Therefore, we can say the length of the rectangle is $11x^{4}$ and width is $(5x^{2} + 1)$ . (Answer)

5 0

3 years ago

In the dryer there are 18 red socks, 24 white socks, 8 blue socks, and 10 black socks. Fred wants to have two matching socks to

liraira [26]

I'm guessing that the answer would be 60

8 0

3 years ago

Q1 A ball is thrown upwards with some initial speed. It goes up to a height of 19.6m and then returns. Find (a) The initial spee

lubasha [3.4K]

Answer:

(a) 19.6 ms⁻¹

(b) 2 s

(d) 4 s

Step-by-step explanation:

<u>Constant Acceleration Equations (SUVAT)</u>

$\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{1}{2}at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^{-1}$\\\\$v$ = final velocity in ms$^{-1}$\\\\$a$ = acceleration in ms$^{-2}$\\\\$t$ = time in s (seconds)\end{minipage}}$

When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.

Acceleration due to gravity = 9.8 ms⁻².

When the ball reaches its maximum height, its velocity will momentarily be zero.

<u>Given values</u> (taking up as positive):

$s=19.6 \quad v=0 \quad a=-9.8$

$\begin{aligned}\textsf{Using} \quad v^2&=u^2+2as\\\\\textsf{Substitute the given values:}\\0^2&=u^2+2(-9.8)(19.6)\\0&=u^2-384.16\\u^2&=384.16\\u&=\sqrt{384.16}\\\implies u&=19.6\; \sf ms^{-1}\end{aligned}$

Therefore, the initial speed is 19.6 ms⁻¹.

Using the same values as for part (a):

$\begin{aligned}\textsf{Using} \quad s&=vt-\dfrac{1}{2}at^2\\\\\textsf{Substitute the given values:}\\19.6&=0(t)-\dfrac{1}{2}(-9.8)t^2\\19.6&=4.9t^2\\t^2&=\dfrac{19.6}{4.9}\\t^2&=4\\t&=\sqrt{4}\\\implies t&=2\; \sf s\end{aligned}$

Therefore, the time taken to reach the highest point is 2 seconds.

As the ball reaches its maximum height at 2 seconds, one second before this time is 1 s.

<u>Given values</u> (taking up as positive):

$u=19.6 \quad a=-9.8 \quad t=1$

$\begin{aligned}\textsf{Using} \quad v&=u+at\\\\\textsf{Substitute the given values:}\\v&=19.6+(-9.8)(1)\\v&=19.6-9.8\\\implies v&=9.8\; \sf ms^{-1}\end{aligned}$

The velocity of the ball one second before it reaches its maximum height is the <u>same</u> as the velocity one second after.

<u>Proof</u>

When the ball reaches its maximum height, its velocity is zero.

Therefore, the values for the downwards journey (from when it reaches its maximum height):

$u=0 \quad a=9.8 \quad t=1$

(acceleration is now positive as we are taking ↓ as positive).

$\begin{aligned}\textsf{Using} \quad v&=u+at\\\\\textsf{Substitute the given values:}\\v&=0+9.8(1)\\\implies v&=9.8\; \sf ms^{-1}\end{aligned}$

Therefore, the velocity of the ball one second before <u>and</u> one second after it reaches the maximum height is 9.8 ms⁻¹.

From part (a) we know that the time taken to reach the highest point is 2 seconds. Therefore, the time taken by the ball to travel from the highest point to its original position will also be 2 seconds.

Therefore, the total time taken by the ball to return to its original position after it is thrown upwards is 4 seconds.

4 0

1 year ago

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