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3 years ago

Which of the following is equivalent to the polynomial below

Mathematics

1 answer:

natta225 [31]3 years ago

8 0

For this case what you must do is find the roots of the corresponding polynomial by means of a factorization.
We have then:
x ^ 2-10x + 61
(x- (5 + 6i)) (x- (5-6i))
Checking:
(x- (5 + 6i)) (x- (5-6i))
x ^ 2 - x * (5-6i) - (5 + 6i) x + (5 + 6i) * (5-6i)
x ^ 2 - 5x + 6ix - 5x- 6ix + 25 - 30i + 30i + 36
x ^ 2 -10x + 61
Answer:
(x- (5 + 6i)) (x- (5-6i))

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A class of 28 students can complete 896 math problems in one day how many problems does each student complete

maksim [4K]

Answer:

32

Step-by-step explanation:

Assuming they all complete it at the same rate,

896/28 = 32

3 0

3 years ago

What conclusions can be made about the amount of money in each account if f represents Molly's account and g represents her brot

Nat2105 [25]

Answer:

(b) is true

Step-by-step explanation:

Given

Molly

$a = 500$ --- starting balance

$m = 10$ --- monthly rate

Her brother

$a = 100$ ---- starting balance

$r = 10\%$ --- annual rate

Required

Determine which option is true

First, we calculate her brother's function.

The function is an exponential function calculated as:

$y = ab^x$

Where $b = 1 + r$

So, we have:

$y = ab^x$

$y = 100 *(1 + 10\%) ^x$

$y = 100 *(1 + 0.10) ^x$

$y = 100 *(1.10) ^x$

Hence:

$g(x) = 100 *(1.10) ^x$

Next, we calculate Molly's function (a linear function)

The monthly function is:

$y = mx + a$

So, we have:

$y = 10x + 500$

Annually, the function will be:

$y = 10x*12 + 500$

$y = 120x + 500$

So, we have:

$f(x) = 120x + 500$

At this point, we have:

$f(x) = 120x + 500$ ---- Molly

$g(x) = 100 *(1.10) ^x$ ---- Her brother

Next, we test each option

(a): Molly's account will have a faster rate of change over [32,40]

We calculated Molly's function to be:

$y = 120x + 500$

The slope of a linear function with the form: $y = mx + b$ is m

By comparison:

$m = 120$

Since Molly's account is a linear function, the rate of change over any interval will always be the same; i.e.

$m = 120$

For his brother:

Rate of change is calculated using:

$m = \frac{g(b) - g(a)}{b - a}$

$m = \frac{g(40) - g(32)}{40 - 32}$

$m = \frac{g(40) - g(32)}{8}$

Calculate g(40) and g(32)

$g(x) = 100 *(1.10) ^x$

$g(40) = 100 * 1.10^{40} =4526$

$g(32) = 100 * 1.10^{32} = 2111$

So, we have:

$m = \frac{4526 - 2111}{8}$

$m = \frac{2415}{8}$

$m = 302$

By comparison: $302 > 120$

Hence, her brother's account has a faster rate over [32,40]

(a) is false

(b): Molly's account will have a slower rate of change over [24,30]

$m = 120$ --- Molly's rate of change

For his brother:

$m = \frac{g(b) - g(a)}{b - a}$

$m = \frac{g(30) - g(24)}{30 - 24}$

$m = \frac{g(30) - g(24)}{6}$

Calculate g(30) and g(24)

$g(x) = 100 *(1.10) ^x$

$g(40) = 100 * 1.10^{30} =1745$

$g(32) = 100 * 1.10^{24} = 985$

So, we have:

$m = \frac{g(30) - g(24)}{6}$

$m = \frac{1745 - 985}{6}$

$m = \frac{760}{6}$

$m = 127$

By comparison: $127 > 120$

Hence, Molly's account has a slower rate over [24,30]

(b) is false

(c): Molly's account will have a slower rate of change over [0,4]

$m = 120$ --- Molly's rate of change

For his brother:

$m = \frac{g(b) - g(a)}{b - a}$

$m = \frac{g(4) - g(0)}{4 - 0}$

$m = \frac{g(4) - g(0)}{4}$

Calculate g(4) and g(0)

$g(x) = 100 *(1.10) ^x$

$g(4) = 100 * 1.10^4 =146$

$g(0) = 100 * 1.10^{0} = 100$

So, we have:

$m = \frac{g(4) - g(0)}{4}$

$m = \frac{146 - 100}{4}$

$m = \frac{46}{4}$

$m = 11.5$

By comparison: $120>11.5$

Hence, Molly's account has a faster rate over [0,4]

(c) is false

4 0

3 years ago

What is -6+-3 equal integers

saul85 [17]

Answer:

Step-by-step explanation:

-6+(-3)

-6-3

Answer is -9

6 0

3 years ago

What is the value of an investment of $650 invested at an annual interest rate of 1.5% compounded monthly for 16 years?

diamong [38]

Answer:

it should be 650

1.5

0.015

Step-by-step explanation:

6 0

2 years ago

Can someone solve this for me?

Helga [31]

well, we know it's a rectangle, so that means the sides JK = IL and JI = KL, so

$\stackrel{JK}{3x+21}~~ = ~~\stackrel{IL}{6y}\implies 3(x+7)=6y\implies x+7=\cfrac{6y}{3} \\\\\\ x+7=2y\implies \boxed{x=2y-7} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{JI}{6y-6}~~ = ~~\stackrel{KL}{2x+20}\implies 6(y-1)=2(x+10)\implies \cfrac{6(y-1)}{2}=x+10 \\\\\\ 3(y-1)=x+10\implies 3y-3=x+10\implies \stackrel{\textit{substituting from the 1st equation}}{3y-3=(2y-7)+10} \\\\\\ 3y-3=2y+3\implies y-3=3\implies \blacksquare~~ y=6 ~~\blacksquare ~\hfill \blacksquare~~ \stackrel{2(6)~~ - ~~7}{x=5} ~~\blacksquare$

5 0

1 year ago