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3 years ago

Identity true statements about quadratic function y = ax^{2} +bx+c, where a =/= 0.

Mathematics

1 answer:

vesna_86 [32]3 years ago

3 0

The true statements are (1) and (2)

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I have to simplify this. It says assume that all variables in the variable in the denominator are nonzero. Leave your answer in

ra1l [238]

I wish I could help u but I can’t

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3 years ago

-3x+18=7x what could you do to isolate the variable term to one side of the equation

Alexxandr [17]

Answer:

7x=−x+24 7 x = − x + 24

Step-by-step explanation:

The equations we solved in the last section simplified nicely so that we could use the. Our strategy will involve choosing one side of the equation to be the variable side, and step by step, to isolate the variable terms on one side of the equation

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2 years ago

Read 2 more answers

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

2 Points

DochEvi [55]

Answer:

Step-by-step explanation:

f(x) = 2x^2 – 3x + 1 is a function (its graph is a parabola that opens up), and it is one-to-one (Answer A). For each input, x, this function will return one output.

8 0

3 years ago

What is the least number of degrees you could rotate a regular pentagon (5 equal side lengths) around its center so that it appe

Leya [2.2K]

Answer:

Well, I'm not the smartest person, nor do I exactly know what you mean, but I have a wild guess that the least you could rotate it is 100 degrees.

Step-by-step explanation:

Considering the fact that it's sides are all equal, you could compare this to a square, which also has equal sides. Its angles are equal, too, so you could measure it's angles (100 degrees) and turn it that exact amount,

I hope this helps you, like I said, I don't exactly know what you are talking about, I am not studying that sort of thing yet.

8 0

3 years ago