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3 years ago

What is the factored or of 25y^4 - 4z^2

Mathematics

1 answer:

Cerrena [4.2K]3 years ago

5 0

Answer:

$(5y^2+2z)(5y^2-2z)$

Step-by-step explanation:

Use the difference of squares formula.

$a^2-b^2=(a+b)(a-b)$

$a=5y^2 \\ b=2z$

$(5y^2+2z)(5y^2-2z)$

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Suppose that a large number of samples are taken from a population. Which of the following sample sizes will be most likely to r

Montano1993 [528]

If you use a large enough statistical sample size, you can apply the Central Limit Theorem (CLT) to a sample proportion for categorical data to find its sampling distribution. The population proportion, p, is the proportion of individuals in the population who have a certain characteristic of interest (for example, the proportion of all Americans who are registered voters, or the proportion of all teenagers who own cellphones). The sample proportion, denoted

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3 years ago

Read 2 more answers

The value of 3b2 – b when b = 5

Mila [183]

Answer:

Step-by-step explanation:

Evaluate 3 b^2 - b where b = 5:

3 b^2 - b = 3×5^2 - 5

Hint: | Evaluate 5^2.

5^2 = 25:

3×25 - 5

Hint: | Multiply 3 and 25 together.

3×25 = 75:

75 - 5

Hint: | Subtract 5 from 75.

| 7 | 5

- | | 5

| 7 | 0:

Answer: 70

6 0

3 years ago

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$

The initial condition equations are now ...

$\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

and the solutions for p and q are ...

$p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}$

Using these formulas on problem (d), we get ...

$c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n$

And for problem (f), we get ...

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0

3 years ago

What is the probability that the spinner lands on an even number or on the unshaded section? one-fifth two-fifths three-fifths f

Tasya [4]

The probability that the spinner lands on an even number or on the unshaded section is; 3/5

<h3>How to find the probability?</h3>

From the spinner attached, we see that the number of sections on the spinner is 5 sections.

Now, we can also see that;

Number of even numbers section = 2

Number of shaded sections = 1

Thus, probability that the spinner lands on an even number or on the unshaded section is;

P(even number or unshaded section) = (2 + 1)/5 = 3/5

Read more about Probability at; brainly.com/question/24756209

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5 0

2 years ago

The area of a square is 4 square inches. what is the length of each side of the square

VladimirAG [237]

2 inches, because 2*2 =4

8 0

3 years ago