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Find the vertices and foci of the hyperbola with equation quantity x minus five squared divided by eighty one minus the quantity

Papessa [141]

If you notice, the positive fraction, is the one with the "x" variable in it, that simply means, the hyperbola is opening over the x-axis, so is a horizontal one, therefore

$\bf \textit{hyperbolas, horizontal traverse axis }
\\\\
\cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1
\qquad 
\begin{cases}
center\ ( h, k)\\
vertices\ ( h\pm a, k)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{ a ^2 + b ^2}
\end{cases}\\\\
-------------------------------$

$\bf \cfrac{(x-5)^2}{81}-\cfrac{(y-1)^2}{144}=1\implies \cfrac{(x-5)^2}{9^2}-\cfrac{(y-1)^2}{12^2}=1~~
\begin{cases}
a=9\\
b=12
\end{cases}
\\\\\\
c=\sqrt{9^2+12^2}\implies c=15\\\\
-------------------------------\\\\
vertices~~(5\pm 9,1)\implies 
\begin{cases}
(14,1)\\
(-4,1)
\end{cases}~~foci~~(5\pm 15,1)\implies 
\begin{cases}
(20,1)\\
(-10,1)
\end{cases}$

3 0

3 years ago

The rate at which a professional tennis player used carbohydrates during a strenuous workout was found to be 1.7 grams per minut

Alex787 [66]

Answer:

m=1.7

C=68 gr

Step-by-step explanation:

Function Modeling

We are given a relationship between the carbohydrates used by a professional tennis player during a strenuous workout and the time in minutes as 1.7 grams per minute. Being C the carbohydrates in grams and t the time in minutes, the model is

$C=1.7t$

The slope m of the line is the coefficient of the independent variable, thus m=1.7

The graph of C vs t is shown in the image below.

To find how many carbohydrates the athlete would use in t=40 min, we plug in the value into the equation

$C=1.7\cdot 40=68\ gr$

8 0

3 years ago

(12h - 54) divided by 6

Stolb23 [73]

12h - 54 : 6 =
(12h - 54)/6 =
6(2h - 9)/6 =
2h - 9

3 0

4 years ago

If a=(1 2 3 4) find A×A and the relation determined by (I) y=2x (II) x+y

kiruha [24]

Answer:

HOPE IT HELPS PLZ MARK ME BRAINLIEST

Step-by-step explanation:

A={1,2,3,4,5,6}

R={(x,y):y is divisible by x}

We know that any number (x) is divisible by itself.

(x,x)∈R

∴R is reflexive.

Now,(2,4)∈R [as 4 is divisible by 2]

But,(4,2)∈ /

R. [as 2 is not divisible by 4]

∴R is not symmetric.

Let (x,y),(y,z)∈R. Then, y is divisible by x and z is divisible by y.

∴z is divisible by x.

⇒(x,z)∈R

∴R is transitive.

Hence, R is reflexive and transitive but not symmetric.

4 0

3 years ago

What is the c-value for the data point that represents the outlier in the scatter plot?

aivan3 [116]

The answer is the first one. 195 is the x value for the data point that represents the outlier in the scatter plot.

8 0

3 years ago