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2 years ago

11

Just list all of the possible answers

2 answers:

Genrish500 [490]2 years ago

8 0

Answer:

what grade are you??........

Ronch [10]2 years ago

4 0

Answer:

8th idea I'm sorry. uhhhhhhhh just wait not

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Solve the following system

scZoUnD [109]

Answer:

{x = -4 , y = 2 , z = 1

Step-by-step explanation:

Solve the following system:

{-2 x + y + 2 z = 12 | (equation 1)

2 x - 4 y + z = -15 | (equation 2)

y + 4 z = 6 | (equation 3)

Add equation 1 to equation 2:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - 3 y + 3 z = -3 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Divide equation 2 by 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Add equation 2 to equation 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+0 y+5 z = 5 | (equation 3)

Divide equation 3 by 5:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 3 from equation 2:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y+0 z = -2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 2 from equation 1:

{-(2 x) + 0 y+2 z = 10 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract 2 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = 8 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = -4 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Collect results:

Answer: {x = -4 , y = 2 , z = 1

4 0

3 years ago

Find the probability of exactly three successes in six trials of a binomial experiment in which the the probability of is 50%.

Lyrx [107]

Start by doing the binomial expansion of (x+y)^6 where x represents success. This is

(x^6y^0) + 6(x^5y^1) +15(x^4y^2) +20(x^3y^3) +15(x^2y^4) +6(x^1y^5) +(x^0y^6)

We are interested in the x^3y^3 term which represents exactly 3 sucesses. Since the probalbility of sucess and failure are both .5 we should be able to figure this out just using the coefficients of the terms which is

20/64 = .3125 which is 31.25%

4 0

3 years ago

A chef uses 4.5 cups of tomatoes to make each batch of pasta sauce. If t represents the number of cups of sauce, which equation

SashulF [63]

Answer: t=4.5x

Step-by-step explanation:

t=cups of tomatoes

x=batch(es) of pasta sauce

Example:

The chef wants to know how many tomatoes he needs to make 2 batches of pasta sauce.

x=2

So 4.5x2=9

Answer for the example is 9

6 0

3 years ago

Read 2 more answers

How do I solve this<br> I’m very confused

nadya68 [22]

First Column, -x and 5 third column, -x and 2

4 0

3 years ago

Read 2 more answers

The polynomial of degree 5, P ( x ) has leading coefficient a=1, has roots of multiplicity 2 at x = 3 and x = 0 , and a root of

ser-zykov [4K]

Answer:

$p_{5} (t) = x^{5} - 5\cdot x^{4} + 3\cdot x^{3} +9\cdot x^{2}$ , for $r_{1} = 0$

Step-by-step explanation:

The general form of quintic-order polynomial is:

$p_{5}(t) = a\cdot x^{5} + b\cdot x^{4} + c\cdot x^{3} + d\cdot x^{2} + e \cdot x + f$

According to the statement of the problem, the polynomial has the following roots:

$p_{5} (t) = (x - r_{1})\cdot (x-3)^{2}\cdot x^{2} \cdot (x+1)$

Then, some algebraic handling is done to expand the polynomial:

$p_{5} (t) = (x - r_{1}) \cdot (x^{3}-6\cdot x^{2}+9\cdot x) \cdot (x+1)\\p_{5} (t) = (x - r_{1}) \cdot (x^{4}-5\cdot x^{3} + 3 \cdot x^{2} + 9 \cdot x)$

$p_{5} (t) = x^{5} - (5+r_{1})\cdot x^{4} + (3 + 5\cdot r_{1})\cdot x^{3} +(9-3\cdot r_{1})\cdot x^{2} - 9 \cdot r_{1}\cdot x$

If $r_{1} = 0$ , then:

$p_{5} (t) = x^{5} - 5\cdot x^{4} + 3\cdot x^{3} +9\cdot x^{2}$

5 0

3 years ago