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Marina86 [1]
3 years ago
10

Someone plzzzz help me):

Mathematics
1 answer:
sergiy2304 [10]3 years ago
3 0

C. x³-4x²-16x+24.

In order to solve this problem we have to use the product of the polynomials where  each monomial of the first polynomial is multiplied by all the monomials that form the second polynomial. Afterwards, the similar monomials are added or subtracted.

Multiply the polynomials (x-6)(x²+2x-4)

Multiply eac monomial of the first polynomial by all the monimials of the second polynomial:

(x)(x²)+x(2x)-(x)(4) - (6)(x²) - (6)(2x) - (6)(-4)

x³+2x²-4x -6x²-12x+24

Ordering the similar monomials:

x³+(2x²-6x²)+(-4x - 12x)+24

Getting as result:

x³-4x²-16x+24

You might be interested in
I just need a little help on this ??! Please.
lawyer [7]

1. Since it's m - 7 you would have 7 to both sides so you would in fact have m < 13. If you double check your answer, you see that if m is say 12 (because 12 is obviously less than 13), 12 - 7 < 6

2. Again, use the same process on this problem as the first one. Add 8 to each sides because it's you're subtracting 8 from n. So you end up with n > 13. Check your answer. Say n is 14. 14 - 8 > 5

3. This one is different because you are adding 5 to p. So in order to get p by itself, you need to subtract 5 from both sides. p < 5. Say p is 4, 4 + 5 < 10.

When working with problems like these, you need to isolate the variable on one side and get it by itself.

5 0
3 years ago
5. If the tree's circumference is 141.3ft, what is its diameter?
BigorU [14]

Answer:

d ≈ 45 ft

General Formulas and Concepts:

<u>Symbols</u>

  • π (pi) ≈ 3.14

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Geometry</u>

  • Circumference of a Circle Formula: C = πd

Step-by-step explanation:

<u>Step 1: Define</u>

Circumference <em>C </em>= 141.3 ft

<u>Step 2: Solve for </u><em><u>d</u></em>

  1. Substitute in <em>C</em> [Circumference Formula]:                                                       141.3 ft = πd
  2. Substitute in π (approximation):                                                                      141.3 ft ≈ 3.14d
  3. [Division Property of Equality] Divide both sides by 3.14:                              45 ft ≈ d
  4. Rewrite:                                                                                                             d ≈ 45 ft
7 0
3 years ago
Read 2 more answers
50,000 contestants participate in an on-line game. The game randomly eliminates 20% of the contestants each day. Determine the n
irga5000 [103]

Answer:10,486 contestants

Step-by-step explanation:

First day:

Total number of contestants at the beginning = 50,000

After first day, 20% 0f 50,000 are removed.

That is, (20/100)×50,000 = 10,000

Contestants remaining after first day = 50,000-10,000 = 40,000

Second day:

Contestants= 40,000

20% removed

That is (20/100)×40000=8000

Balance =40000-8000=32,000

Third day:

Contestants= 32000

20% removed

That is (20/100)×32000=6400

Balance =32000-6400=25,600

Fouth day:

Contestants= 25,600

20% removed

That is (20/100)×25,600=5,120

Balance =25,600-5,120=20,480

Fifth day:

Contestants= 20,480

20% removed

That is (20/100)×20,480=4096

Balance =20,480-4096=16,384

Sixth day:

Contestants= 16,384

20% removed

That is (20/100)×16,384=3276.8

Balance =16,384-3276.8=13,107.2

Seventh day:

Contestants= 13,107.2

20% removed

That is (20/100)×13,107.2=2621.44

Balance =13107.2-2621.44=10485.76

Therefore, number of contestants remaining after one week equals 10,486.

6 0
3 years ago
I need help with this please
Evgesh-ka [11]
Answer:

3) 0 ≤ y ≤ 100
7 0
3 years ago
Show work please.<br><br> solve system of equations using matrices.
nadya68 [22]

Answer:

(t, t -1, t)

Step-by-step explanation:

You have three unknowns but only 2 equations, so you can't really SOLVE this...you can get a solution with a variable still in it (I forget what this is called.  I think it refers to infinite many solutions).  Here's how it works:

Set up your matrix:

\left[\begin{array}{ccc}1&-2&1\\2&-1&-1\\\end{array}\right] \left[\begin{array}{ccc}2\\1\\\end{array}\right]

You want to change the number in position 21 (the 2 in the scond row) to a 0 so you have y and z left.  Do this by multiplying the top row by -2 then adding it to the second row to get that 2 to become a 0.  Multiplying in a -2 to the top row gives you:

\left[\begin{array}{ccc}-2&4&-2\\2&-1&-1\\\end{array}\right]\left[\begin{array}{ccc}-4\\1\\\end{array}\right]

Then add, keeping the first row the same and changing the second to reflect the addition:

\left[\begin{array}{ccc}-2&4&-2\\0&3&-3\\\end{array}\right] \left[\begin{array}{ccc}-4\\-3\\\end{array}\right]

The second equation is this now:

3y - 3z = -3.  Solving for y gives you y = z - 1.  Let's let z = t (some random real number that will make the system true.  Any number will work.  I'll show you at the end.  Just bear with me...)

lf z = t, and if y = z - 1, then y = t - 1.  So far we have that y = t - 1 and z = t.  Now we solve for x:

From the first equation in the original system,

x - 2y + z = 2.  Subbing in t - 1 for y and t for z:

x - 2(t - 1) + t = 2.  Simplify to get

x - 2t + 2 + t = 2  and  x - t = 0, and x = t.  So the solution set is (t, t - 1, t).  Picking a random value for t of, let's say 2, sub that in and make sure it works.  If:

x - 2y + z = 2, then t - 2(t - 1) + t = 2 becomes t - 2t + 2 + t = 2, and with t = 2, 2 - 2(2) + 2 + 2 = 2.    Check it:  2 - 4 + 4 = 2 and 2 = 2.  You could pick any value for t and it will work.

6 0
2 years ago
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