What is the value of 2 in 0.259

How can sums and differences of cubes be identified for factoring?

Ad libitum [116K]

The sum of cubes is given as:
a³ + b³ = (a + b)(a² - ab + b²)

Example for the sum of cubes:

64x³+y³ ⇒ This is the sum of cubes because each term; 64, x³, and y³ are cube numbers

By writing each term as an expression of cube numbers, we have:
(4x)³ + (y)³ ⇒ 64 is 4³
Use the factorization of the sum of cubes, we have:
(4x + y) ( (4x)²- 4xy + y²)
(4x + y) (16x² - 4xy + y²)

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The difference of cubes can be factorized as:
(x³ - y³) = (x - y)(x² + xy + y²)

Example
(125x³ - 8y³) = (5x - 2y) ((5x)² + (5x)(2y) + (2y)²)
= (5x - 2y) (25x² + 10xy + 4y²)

7 0

3 years ago

QUESTION 8 The orthocenter of a triangle is formed by the ____________of a triangle. perpendicular bisectors angle bisectors med

Otrada [13]

<span>The orthocenter of a triangle is formed by the ALTITUDES of a triangle.</span>

Source:
http://www.1728.org/trictr.htm

4 0

2 years ago

What is the place value 8 in 880,000

polet [3.4K]

Answer:

hundred thousands

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

A well known pharmaceutical manufacturer is manufacturing a newly developed vaccine, and is concerned about variability in the i

Lana71 [14]

Answer:

a

The null hypothesis is $H_o : \ \sigma^2 = 1.9$

The alternative hypothesis is $H_a : \ \sigma^2 \ne 1.9$

b

$X^2 = 13.74$

c

The decision rule is

Reject the null hypothesis

d

The conclusion is

There is no sufficient evidence to conclude that the variance of the immune response is equal to 1.9.

Step-by-step explanation:

From the question we are are told that

The variance is $\sigma ^2 = 1.9$

The sample size is n = 30

The sample variance is $s^2 = 0.9$

The level of significance is $\alpha = 0.05$

The null hypothesis is $H_o : \ \sigma^2 = 1.9$

The alternative hypothesis is $H_a : \ \sigma^2 \ne 1.9$

Generally the test statistics is mathematically represented as

$X^2 = \frac{ (n- 1 ) * s^2 }{ \sigma^2 }$

=> $X^2 = \frac{ (30 - 1 ) * 0.9 }{1.9 }$

=> $X^2 = 13.74$

Generally from the degree of freedom is mathematically represented as

$df = n- 1$

=> $df = 30 - 1$

=> $df = 29$

Generally from the chi distribution table the critical value of $\frac{\alpha}{2} \ and \ 1 - \frac{ \alpha }{2}$ at a degree of freedom of $df = 29$ is