True or false: The roots of a certain quartic equations are +-1, 0, and 1+i. (All equations mentioned have real coefficients)

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

3 0

3 years ago

A bag contains a variety of different colored marbles. P( red)= 1/2, P( green) = 1/4, and P( red and green) = 1/8 , which statem

Veronika [31]

Answer:

Step-by-step explanation:

The 1 choice. The events are independent because P(red) ×P(green)= P(red and green)

6 0

3 years ago

Read 2 more answers

Which story represents the graph below?

-BARSIC- [3]

Answer:

c. both

Step-by-step explanation:

6 0

3 years ago

Which equation has a constant of proportionality equal to 9 y=81/3x y=9x y=3x y==1/9x

aev [14]

Answer:

B) y = 9 x has Proportionality Constant = 9

Step-by-step explanation:

Two quantities P and Q are said to be PROPORTIONAL if and only if:

P ∝ Q ⇔ P = k Q ⇔ $k = \frac{P}{Q}$

Here, k = PROPORTIONALITY CONSTANT

Given : k = 9

Now, let us consider the given expressions in which x ∝y

y = 81/3 x

Here, $\frac{y}{x} = \frac{81}{3} = 27 \ne 9$

So, here Proportionality Constant ≠ 9

y = 9 x

Here, $\frac{y}{x} = 9$

So, here Proportionality Constant = 9

y = 3 x

Here, $\frac{y}{x} = 3 \ne 9$

So, here Proportionality Constant ≠ 9

y = 1/9 x

Here, $\frac{y}{x} = \frac{1}{9} \ne 9$

So, here Proportionality Constant ≠ 9

Hence, only y = 9 x has Proportionality Constant = 9

8 0

3 years ago

What are the zeros of the quadratic function? f(x)=3x2−300 Enter your answers in the boxes.

nata0808 [166]

Hi there!

We can find the zeros of a function by solving f(x) = 0. We find the following:

$3 {x}^{2} - 300 = 0$
Add 300 to both sides.

$3 {x}^{2} = 300$
Divide by 3.

${x}^{2} = 100$
Take the root and minus root of both sides.

$x = \sqrt{100} = 10 \\ or \\ x = - \sqrt{100} = - 10$

Therefore, the zeros are
x = 10 and x = -10.

~ Hope this helps you!

5 0

3 years ago