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2 years ago

I NEED HELP PLEASE and thx

Mathematics

1 answer:

Galina-37 [17]2 years ago

3 0

Answer:

Step-by-step explanation:

By multiplying -2x by 3 to get -6x, which will cancel out with 6x.

b. We could also multiply -4y by -2 to get 8y, so it cancels out with -8y.

Hope this helps!

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3 years ago

Read 2 more answers

If a polynomial has one root in the form a+√b , it has a second root in the form of a_√b

gulaghasi [49]

When roots of polynomials occur in radical form, they occur as two conjugates.
That is,
The conjugate of (a + √b) is (a - √b) and vice versa.
To show that the given conjugates come from a polynomial, we should create the polynomial from the given factors.

The first factor is x - (a + √b).
The second factor is x - (a - √b).

The polynomial is
f(x) = [x - (a + √b)]*[x - (a - √b)]
= x² - x(a - √b) - x(a + √b) + (a + √b)(a - √b)
= x² - 2ax + x√b - x√b + a² - b
= x² - 2ax + a² - b

This is a quadratic polynomial, as expected.

If you solve the quadratic equation x² - 2ax + a² - b = 0 with the quadratic formula, it should yield the pair of conjugate radical roots.
x = (1/2) [ 2a +/- √(4a² - 4(a² - b)]
= a +/- (1/2)*√(4b)
= a +/- √b
x = a + √b, or x = a - √b, as expected.

3 0

3 years ago

Read 2 more answers

you measure the period of a mass oscillating on a vertical spring ten times as follows: period (s): 1.06, 1.31, 1.28, 0.99, 1.48

lidiya [134]

The mean and (sample) standard deviation σ = 0.2098.

<h3>What exactly would the standard deviation indicate?</h3>

The term "standard deviation" (or "") refers to the degree of dispersion of the data from the mean. Data are grouped around the mean when the standard deviation is low, and are more dispersed when the standard deviation is high.

<h3>According to given information:</h3>

The mean is the product of the dataset's total and the sample size. Mathematically.

$\bar{x}=\frac{\sum X_i}{N}$

The individual periods are Xi.

The sample size is N.

$\sum X i$ = 1.06 + 1.31 + 1.28 + 0.99,+ 1.48 + 1.37+ 0.98 + 1.31 + 1.59 + 1.55

$\sum X i$ = 12.92

N = 10

While substituting the value we get:

x = 12.96/10

x = 1.292

The samples' average is 1.292.

The standard deviation:

$\sigma=\sqrt{\frac{\sum(x-\bar{x})^2}{N}}$

$\sum(x-\bar{x})^2$ = (1.48-1.292)^2+(1.37-1.292)^2+(0.98-1.292)^2+(1.31-1.292)^2+(1.59-1.292)^2+(1.55-1.292)^2.

$\sum(x-\bar{x})^2$ = 0.43996

Putting into the formula we get:

$\sigma=\sqrt{\frac{0.43996}{10}}$

σ = √(0.043996)

σ = 0.2098

The mean and (sample) standard deviation σ = 0.2098.

To know more about standard deviation visit:

brainly.com/question/18521100

#SPJ4

I understand that the question you are looking for is:

You measure the period of a mass oscillating on a vertical spring ten times as follows:

Period (s): 1.06, 1.31, 1.28, 0.99, 1.48, 1.37, 0.98, 1.31, 1.59, 1.55

Required:

What are the mean and (sample) standard deviation?

a. Mean: 1.228, Standard Deviation: 0.2135

b. Mean: 1.325, Standard Deviation: 0.1674

c. Mean: 1.292. Standard Deviation: 0.2211

d. Mean: 1.228, Standard Deviation: 0.2098

e. Mean: 1.292, Standard Deviation: 0.2135

7 0

2 years ago

I NEED HELP PLEASE and thx ​

I NEED HELP PLEASE and thx