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3 years ago

The hyperbola with center at (4, 5), with vertices at (0, 5) and (8, 5), with b = 2.

1 answer:

galina1969 [7]3 years ago

8 0

The standard form of hyperbola is:
(x-h)²/a²-(y-k)²/b²=1
center:(4,5)
Length of the horizontal transverse axis: 8-5=3=2a
thus
a=3/2
a²=9/4
b=2
b²=4

Hence the equation will be:
(x-4)²/(9/4)-(y-5)²/4=1
simplifying this we get:
[4(x-4)²]/9-(y-5)²/4=1

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Find all the zeros of the polynomial function p(x) = x3 – 5x2 + 33x – 29

hram777 [196]

Answer:

$\large \boxed{\sf \ \ x=1, \ \ x=2+5i, \ \ x=2-5i \ \ }$

Step-by-step explanation:

Hello,

I assume that we are working in $\mathbb{C}$ , otherwise there is only one zero which is 1. Please consider the following.

First of all, <u>we can notice that 1 is a trivial solution</u> as

$p(1) = 1^3-5\cdot 1^2 + 33\cdot 1-29=1-5+33-29=0$

It means that (x-1) is a factor of p(x) so we can find two real numbers, a and b, so that we can write the following.

$p(x)=(x-1)(x^2+ax+b)=x^3+ax^2+bx-x^2-ax-b=x^3+(a-1)x^2+(b-a)x-b$

Let's identify like terms as below.

a-1 = -5 <=> a = -5 + 1 = -4

b-a = 33

-b = -29 <=> b = 29

$\boxed{ \ p(x)=(x-1)(x^2-4x+29) \ }$

Now, we need to find the zeroes of the second factor, meaning finding x so that:

$x^2-4x+29=0 \ \text{ complete the square, 29 = 25 + 4} \\ \\ x^2-2\cdot 2 \cdot x+2^2+25=0 \\ \\ (x-2)^2=-25=(5i)^2 \ \text{ take the root } \\ \\x-2=\pm 5i \ \text{ add 2 } \\ \\ x = 2+5i \ \text{ or } \ x = 2-5i$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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ANSWER PLEASE Tyler deposits $2000 and has a 8% interest compound quarterly. How much does he have Quarter 1,2,3,4 Show Work

guajiro [1.7K]

Answer:

If the time passed is only 3 months, then it is $2040

Step-by-step explanation:

We can use the quarterly compounded interest equation for this problem: P(1 + r/n)^nt

Step 1: Find out how much 3 months is in a year

<em>In this case, 3/12 which is 1/4</em>

Step 2: Plug in known variables into equation

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