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3 years ago

Can someone please help to write the standard form of this hyperbola?

Mathematics

1 answer:

baherus [9]3 years ago

5 0

Hyperbole is the sound of something. For instance: The bees went buzzing by. Buzzing is hyperbole.

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(-4, -2),(-18,7) slope

Nuetrik [128]

Slope = (7 + 2)/(-18 + 4) = 9/-14= - 9/14

3 0

3 years ago

Is the number <img src="https://tex.z-dn.net/?f=%5Csqrt%7B62%7D" id="TexFormula1" title="\sqrt{62}" alt="\sqrt{62}" align="absmi

mrs_skeptik [129]

Answer:

Irrational.

Step-by-step explanation:

Rational means that it is a whole number and can be written as a fraction. That doesn't apply for square rooted numbers. So, $\sqrt{62}$ is irrational.

4 0

4 years ago

The water dam is filled up using the first tributary in 1 hour and 10 minutes, using the second

kondaur [170]

Answer:

I would use quaro is is faster.

Step-by-step explanation:

7 0

2 years ago

How do you find a vector that is orthogonal to 5i + 12j ?

Rashid [163]

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\
slope=\cfrac{a}{{{ b}}}\qquad negative\implies -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\
-------------------------------\\\\$

$\bf \boxed{5i+12j}\implies 
\begin{array}{rllll}
\ \textless \ 5&,&12\ \textgreater \ \\
x&&y
\end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5}
\\\\\\
slope=\cfrac{12}{{{ 5}}}\qquad negative\implies -\cfrac{12}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{12}
\\\\\\
\ \textless \ 12, -5\ \textgreater \ \ or\ \ \textless \ -12,5\ \textgreater \ \implies \boxed{12i-5j\ or\ -12i+5j}$

if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.

so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.

so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.

or using a unit vector for those above, then

$\bf \textit{unit vector}\qquad \cfrac{\ \textless \ a,b\ \textgreater \ }{||\ \textless \ a,b\ \textgreater \ ||}\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{\sqrt{a^2+b^2}}\implies \cfrac{a}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}
\\\\\\
\cfrac{12,-5}{\sqrt{12^2+5^2}}\implies \cfrac{12,-5}{13}\implies \boxed{\cfrac{12}{13}\ ,\ \cfrac{-5}{13}}
\\\\\\
\cfrac{-12,5}{\sqrt{12^2+5^2}}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}$

4 0

4 years ago

Suppose the time required for an auto shop to do a tune-up is normally distributed, with a mean of 102 minutes and a standard de

hammer [34]

Answer:

Step-by-step explanation:

Suppose the time required for an auto shop to do a tune-up is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - u)/s

Where

x = points scored by students

u = mean time

s = standard deviation

From the information given,

u = 102 minutes

s = 18 minutes

1) We want to find the probability that a tune-up will take more than 2hrs. It is expressed as

P(x > 120 minutes) = 1 - P(x ≤ 120)

For x = 120

z = (120 - 102)/18 = 1

Looking at the normal distribution table, the probability corresponding to the z score is 0.8413

P(x > 120) = 1 - 0.8413 = 0.1587

2) We want to find the probability that a tune-up will take lesser than 66 minutes. It is expressed as

P(x < 66 minutes)

For x = 66

z = (66 - 102)/18 = - 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.02275

P(x < 66 minutes) = 0.02275

4 0

4 years ago