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2 years ago

1. A recliner that costs $150 was marked up by 100%.

Mathematics

1 answer:

geniusboy [140]2 years ago

3 0

1. 150 markup and 300 selling

2. 19.2 markup and 67.2 selling

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Which choice is equivalent to the expression below? 12^-2<br> 12 Points!!!!

Brut [27]

Step-by-step explanation:

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2 years ago

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Someone please help me

Ratling [72]

Answer:

1 mi/1.61 km

Step-by-step explanation:

we know that

1 mile= 1.61 kilometers

To convert 1 kilometer to miles

use proportion

1/1.61 mi/km=x/1 mi/km

x=1 mi/1.61 km

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2 years ago

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The credit remaining on a phone card (in dollars) is a linear function of the total calling time made with the card (in minutes)

saw5 [17]

Answer: $19.6

Step-by-step explanation:

Linear function: f(x)=mx+c

, where m= rate of change in f(x) with respect to x.

c = Initial value.

Let c = Initial value of card , m= Charge per minute

x= Number of minutes calling time.

Then, 25.06= 38m+c (i)

21.03=69m+c (ii)

Eliminate (ii) from (i)

$4.03=-31m\\\\\Rightarrow\ m=\dfrac{4.03}{-31}\\\\\Rightarrow\ m=-0.13$

Put m in (i) , we get

$25.06= 38(-0.13)+c\\\\\Rightarrow\ 25.06=-4.94+c\\\\\Rightarrow\ c=25.06+4.94\\\\\Rightarrow\ c=30$

i.e. f(x)=-0.13x+30

if x=80 then

f(80)= -0.13(80)+30

=-10.4+30

=19.6

Hence, the remaining credit after 80 minutes of calls = $19.6

7 0

3 years ago

Marine biologists have determined that when a shark detectsthe presence of blood in the water, it will swim in the directionin w

siniylev [52]

Solution :

a). The level curves of the function :

$C(x,y) = e^{-(x^2+2y^2)/10^4}$

are actually the curves

$e^{-(x^2+2y^2)/10^4}=k$

where k is a positive constant.

The equation is equivalent to

$x^2+2y^2=K$

$\Rightarrow \frac{x^2}{(\sqrt K)^2}+\frac{y^2}{(\sqrt {K/2})^2}=1, \text{ where}\ K = -10^4 \ln k$

which is a family of ellipses.

We sketch the level curves for K =1,2,3 and 4.

If the shark always swim in the direction of maximum increase of blood concentration, its direction at any point would coincide with the gradient vector.

Then we know the shark's path is perpendicular to the level curves it intersects.

b). We have :

$\triangledown C= \frac{\partial C}{\partial x}i+\frac{\partial C}{\partial y}j$

$\Rightarrow \triangledown C =-\frac{2}{10^4}e^{-(x^2+2y^2)/10^4}(xi+2yj),$ and

$\triangledown C$ points in the direction of most rapid increase in concentration, which means $\triangledown C$ is tangent to the most rapid increase curve.