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2 years ago

Maximum and Minimum Values of General Quadratic Functions

Mathematics

1 answer:

e-lub [12.9K]2 years ago

7 0

Answer:

x=-1

y=15

Step-by-step explanation:

This works because completing the square is essentially demonstrating what g(x) - where g(x) in this case = x^2 - would have to be translated by to give the function you have.

So by completing the square you have shown that the function you've been given has been shifted to the left by -1, stretched by a factor of 2 along the y axis and shifted up by 15.

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Pls look at photos I’ll give 30 points

puteri [66]

You said you’d give 30 points but i only see 15 there ☹️

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2 years ago

Need help with number 2 and 3

IceJOKER [234]

Answer:

Yes, you can square a negative number. In fact, any number at all can be squared, even numbers like pi and 0. This is because to square a number just means to multiply it by itself. ... Note that this is positive because when you multiply two negative numbers you get a positive result.

Step-by-step explanation:

number 2

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2 years ago

PLS HELP

Gnesinka [82]

Answer:

Step-by-step explanation:

Let the required number be 100.

Now, according to the question, the number is first increased by 30 %

⇒ 100 + 30 % of 100

⇒ 100 + [(30*100)/100]

⇒ 100 + 30

⇒ 130

Then, it is decreased by 30 %

⇒ 130 - (30 % of 130)

⇒ 130 - [(30*130)/100]

⇒ 130 - 39

⇒ 91

Net Decrease = 100 - 91

= 9

Percent Decrease = (Change in number*100)/100

⇒ (9*100)/100

= 9 %

6 0

3 years ago

What is the binomial expansion of (x + 2)4? x4 + 4x3 + 6x2 + 4x + 1 8x3 + 24x2 + 32x x4 + 8x3 + 24x2 + 32x + 16 2x4 + 8x3 + 12x2

Margaret [11]

<u>Answer-</u>

$\boxed{\boxed{(x+2)^4=x^4+8x^3+24x^2+32x+16}}$

<u>Solution-</u>

Given expression is $(x+2)^4$

Applying Binomial Theorem

$\left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

Here,

a = x, b = 2 and n = 4

So,

$\left(x+2\right)^4=\sum _{i=0}^4\binom{4}{i}x^{\left(4-i\right)}\cdot \:2^i$

Expanding the summation

$=\dfrac{4!}{0!\left(4-0\right)!}x^4\cdot \:2^0+\dfrac{4!}{1!\left(4-1\right)!}x^3\cdot \:2^1+\dfrac{4!}{2!\left(4-2\right)!}x^2\cdot \:2^2+\dfrac{4!}{3!\left(4-3\right)!}x^1\cdot \:2^3+\dfrac{4!}{4!\left(4-4\right)!}x^0\cdot \:2^4$

$=\dfrac{4!}{0!\left(4\right)!}x^4\cdot \:2^0+\dfrac{4!}{1!\left(3\right)!}x^3\cdot \:2^1+\dfrac{4!}{2!\left(2\right)!}x^2\cdot \:2^2+\dfrac{4!}{3!\left(1\right)!}x^1\cdot \:2^3+\dfrac{4!}{4!\left(0\right)!}x^0\cdot \:2^4$