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NISA [10]
3 years ago
6

What are the slopes of the asymptotes of a hyperbola with equation (y-2)^2/1^2 - (x-4)^2/3^2 = 1?

Mathematics
1 answer:
vladimir1956 [14]3 years ago
8 0

Answer: 1/3 and -1/3

Step-by-step explanation:

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Which choice is the explicit formula for the following geometric sequence?0.5, -0.1, 0.02, -0.004, 0.0008, ...
Minchanka [31]
a_1=0.5\\a_2=-0.1\\a_3=0.02\\a_4=-0.004\\a_5=0.0008\\\\a_n=a_1\cdot q^{n-1}\\\\q=a_2:a_1\to q=-0.1:0.5=-0.2\\\\a_n=0.5\cdot(-0.2)^{n-1}=0.5\cdot(-0,2)^n\cdot(-0.2)^{-1}\\\\=0.5\cdot(-0.2)^n\cdot(-5)=-2.5\cdot(-0.2)^n\\\\\boxed{a_n=-2.5\cdot(-0.2)^n}
6 0
3 years ago
A different whole number from 1 - 26 is assigned to each letter of the alphabet and written on the kindergarten 26 alphabet bloc
elena55 [62]
The sum of these will simply be the sum of all the numbers 1 through 26, which is 351.
8 0
3 years ago
How do I solve an equation like this, one with a negative in front of parentheses?
azamat

The factor outside the parentheses is -1. Distribute using that factor.

2 - (4 - <em>x</em>) = 7<em>x</em> - 5<em>x</em>

<em />

-1 * 4 = -4 and -1 * -<em>x</em> = <em>x</em>

<em />

2 - 4 + <em>x</em> = 7<em>x</em> - 5<em>x</em>

Simplify.

-2 + <em>x</em> = 2<em>x</em>

<em>x</em> = 2<em>x</em> + 2

-<em>x</em> = 2

<em>x</em> = -2

<h3>Answer:</h3>

<em>x</em> = -2

3 0
3 years ago
Three numbers have a sum of 82. The second number is twice the first number. The third number is 7 more than the second number.
iVinArrow [24]

Answer: The number is 15

Step-by-step explanation: basically the equation you should solve is

(x)+(2x)+(2x+7)=82

5x+7=82

5x= -7+82

5x=75

5/5x=75/5

x=15

3 0
3 years ago
2. Margie Spencer wants to have $100,000 in her savings account in 20 years. If her account pays 6.6% annual
WITCHER [35]

~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \$100000\\ P=\textit{original amount deposited}\\ r=rate\to 6.6\%\to \frac{6.6}{100}\dotfill &0.066\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{semiannually, thus two} \end{array}\dotfill &2\\ t=years\dotfill &20 \end{cases}

100000=P\left(1+\frac{0.066}{2}\right)^{2\cdot 20}\implies 100000=P(1.033)^{40} \\\\\\ \cfrac{100000}{1.033^{40}}=P\implies 27288.97\approx P

7 0
2 years ago
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