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SashulF [63]
2 years ago
10

How do i express x^2-3x into the form of (x-m)^2+n. I will give brainliest to correct answer

Mathematics
1 answer:
mel-nik [20]2 years ago
7 0

Step-by-step explanation:

let's simply do the multiplications and then compare with the original.

(x-m)² + n

right ?

or is it

{(x - m)}^{2 + n}

let's go for the first.

x² - 2mx + m² + n = x² - 3x

-2mx + m² + n = -3x

fun there we see two things :

-2m = -3

m = 3/2

and

m² + n = 0

(3/2)² = -n

9/4 = -n

n = -9/4

so our transformed expression looks like

(x - 3/2)² - 9/4

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9 is the geometric mean between 4 and which other number?
KonstantinChe [14]
Answer is A. 20.25

cause 20.25 x 9 = 81
√81 = 9
so 9 is the geometric mean <span>between 4 and 20.25</span>
5 0
3 years ago
The regular price of a shirt is $15. Natalie bought it on sale for 0.9 times its regular price. Then the sale price was multipli
bezimeni [28]

Answer: $14.31

Step-by-step explanation:

Assuming it is the final price Natalie paid that you want:

Selling price = 0.9 * 15

= $13.50

To cater for tax it was then multiplied by 1.06:

= 13.5 * 1.06

= $14.31

6 0
3 years ago
(3*10)*8=3*(10*8) what property is this
Dahasolnce [82]
That is the Community Property of Multiplication.

Hope that helps :)
8 0
3 years ago
Read 2 more answers
Given that 'n' is any natural numbers greater than or equal 2. Prove the following Inequality with Mathematical Induction
Oliga [24]

The base case is the claim that

\dfrac11 + \dfrac12 > \dfrac{2\cdot2}{2+1}

which reduces to

\dfrac32 > \dfrac43 \implies \dfrac46 > \dfrac86

which is true.

Assume that the inequality holds for <em>n</em> = <em>k </em>; that

\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k > \dfrac{2k}{k+1}

We want to show if this is true, then the equality also holds for <em>n</em> = <em>k</em> + 1 ; that

\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2(k+1)}{k+2}

By the induction hypothesis,

\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2k}{k+1} + \dfrac1{k+1} = \dfrac{2k+1}{k+1}

Now compare this to the upper bound we seek:

\dfrac{2k+1}{k+1}  > \dfrac{2k+2}{k+2}

because

(2k+1)(k+2) > (2k+2)(k+1)

in turn because

2k^2 + 5k + 2 > 2k^2 + 4k + 2 \iff k > 0

6 0
2 years ago
Read 2 more answers
The sample size needed to estimate the difference between two population proportions to within a margin of error E with a signif
devlian [24]

Answer:

n= (z)22E2

n=10× 99%÷ 0.07

3 0
3 years ago
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