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3 years ago

Write this polynomial in standard form :8x^2 - 4x^3 + 12 - 5x

Mathematics

1 answer:

rodikova [14]3 years ago

3 0

Hello from MrBillDoesMath

Answer: -4x^3 + 8x^2 -5x + 12

Discussion:

Standard for for a polynomial meas tne power of x descend so standard fro of

8x^2 - 4x^3 +12 -5x is

-4x^3 + 8x^2 -5x + 12

Note: the problem statements shows this ":8x^2". I didn't know how to deal with the ":" character so simply ignored it.

Thank you,

Mr. B

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Hi guys, Can anyone help me with this tripple integral? Thank you:)

OleMash [197]

I don't usually do calculus on Brainly and I'm pretty rusty but this looked interesting.

We have to turn K into the limits of integration on our integrals.

Clearly 0 is the lower limit for all three of x, y and z.

Now we have to incorporate

x+y+z ≤ 1

Let's do the outer integral over x. It can go the full range from 0 to 1 without violating the constraint. So the upper limit on the outer integral is 1.

Next integral is over y. y ≤ 1-x-z. We haven't worried about z yet; we have to conservatively consider it zero here for the full range of y. So the upper limit on the middle integration is 1-x, the maximum possible value of y given x.

Similarly the inner integral goes from z=0 to z=1-x-y

We've transformed our integral into the more tractable

$\displaystyle \int_0^1 \int_0^{1-x} \int _0^{1-x-y} (x^2-z^2)dz \; dy \; dx$

For the inner integral we get to treat x like a constant.

$\displaystyle \int _0^{1-x-y} (x^2-z^2)dz = (x^2z - z^3/3)\bigg|_{z=0}^{z= 1-x-y}=x^2(1-x-y) - (1-x-y)^3/3$

Let's expand that as a polynomial in y for the next integration,

$= y^3/3 +(x-1) y^2 + (2x+1)y -(2x^3+1)/3$

The middle integration is

$\displaystyle \int_0^{1-x} ( y^3/3 +(x-1) y^2 + (2x+1)y -(2x^3+1)/3)dy$

$= y^4/12 + (x-1)y^3/3+ (2x+1)y^2/2- (2x^3+1)y/3 \bigg|_{y=0}^{y=1-x}$

$= (1-x)^4/12 + (x-1)(1-x)^3/3+ (2x+1)(1-x)^2/2- (2x^3+1)(1-x)/3$

Expanding, that's

$=\frac{1}{12}(5 x^4 + 16 x^3 - 36 x^2 + 16 x - 1)$

so our outer integral is

$\displaystyle \int_0^1 \frac{1}{12}(5 x^4 + 16 x^3 - 36 x^2 + 16 x - 1) dx$

That one's easy enough that we can skip some steps; we'll integrate and plug in x=1 at the same time for our answer (the x=0 part doesn't contribute).

$= (5/5 + 16/4 - 36/3 + 16/2 - 1)/12$

$=0$

That's a surprise. You might want to check it.

Answer: 0

6 0

3 years ago

Given the event "a die lands with a 6 on top", which of the following is the complement of this event? Choose the correct ans

Veronika [31]

Answer:

The die lands with a 1, 2, 3, 4, or 5 on top ⇒ answer D

Step-by-step explanation:

- Complementary events are two outcomes of an event that are the

only two possible outcomes

- The Complement Rule states that the sum of the probabilities of

an event and its complement must equal 1

- P(A) + P(A') = 1

- Lets solve the problem

- The event A is:

" a die lands with a 6 on top "

∴ P(A) = $\frac{1}{6}$

∵ P(A) + P(A') = 1

∴ $\frac{1}{6}$ + P(A') = 1

- Subtract $\frac{1}{6}$ from both sides

∴ P(A') = $\frac{5}{6}$

∵ The die has six numbers from 1 to 6

∴ The other numbers than 6 are 1 , 2 , 3 , 4 and 5

∴ The complement of event A is:

" A die lands with 1 , 2 , 3 , 4 or 5 "

* The complement of the event is:

The die lands with a 1, 2, 3, 4, or 5 on top

8 0

3 years ago

Simplify 7(x - 1) - 2(x + 1) + 3(x - 4) 8x - 21 8x - 4 8x + 21

Anna007 [38]

Answer:

all work is shown and pictured

7 0

2 years ago

Timothy is filling fish bowls with pebbles at his fish store. He wants to have $\frac{1}{6}$ of a pound of pebbles in each fish

Luba_88 [7]

Answer:

Step-by-step explanation:

Given that:

Each fish bowl to have pebbles of weight equivalent to = $\frac{1}{6}$

Total pounds of pebbles that Timothy can use = $2\frac{1}{3}$

To find:

The greatest value of Total number of fish bowls that Timothy can fill ?

Solution:

First of all, we need to convert mixed fraction into a fractional number and then we also need to see division of two fractions.

Formula:

$1. \ p\dfrac{q}{r} = \dfrac{p\times r+q}{r}\\2. \ \dfrac{\frac{a}{b}}{\frac{c}d}=\dfrac{a\times d}{b\times c}$

Now, the given mixed fraction can be converted to fractional number as:

$2\dfrac{1}{3} = \dfrac{2\times 3+1}{3} = \dfrac{7}{3}$

Now, To find the total number of fish bowls that can be filled, we need to divide the total number of pounds with number of pounds of pebbles in each fish bowl.

So, the answer is:

$\dfrac{\frac{7}{3}}{\frac{1}{6}}\\\Rightarrow \dfrac{7\times 6}{3\times 1}\\\Rightarrow \dfrac{42}{3}\\\Rightarrow \bold{14}$

14 number of fish bowls can be filled.

3 0

3 years ago

Does anyone know these answers please help

Lyrx [107]

1) this is simply solving it, so if you divide 16 by 3, you end up getting 5 1/3

2) 3 1/2, by multiplying 7 by 1/2

3) 224, by multiplying 14 by 16

4) 54 mph, by dividing 324 by 6

5) blue, because only the 1/4 ribbon color would be collected after 3/4 mile

6) 7/8 - 3/8 equals 4/8, or 1/2 when simplified

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2 years ago