ILL MARK BRAINIEST IF YOU DO THIS RIGHT!!! DO 9 AND 10!!!

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

A cylinder shaped can needs to be constructed to hold 400 cubic centimeters of soup. The material for the sides of the can costs

LenKa [72]

Answer:

The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.

Step-by-step explanation:

Volume of the Cylinder=400 cm³

Volume of a Cylinder=πr²h

Therefore: πr²h=400

$h=\frac{400}{\pi r^2}$

Total Surface Area of a Cylinder=2πr²+2πrh

Cost of the materials for the Top and Bottom=0.06 cents per square centimeter

Cost of the materials for the sides=0.03 cents per square centimeter

Cost of the Cylinder=0.06(2πr²)+0.03(2πrh)

C=0.12πr²+0.06πrh

Recall: $h=\frac{400}{\pi r^2}$

Therefore:

$C(r)=0.12\pi r^2+0.06 \pi r(\frac{400}{\pi r^2})$

$C(r)=0.12\pi r^2+\frac{24}{r}$

$C(r)=\frac{0.12\pi r^3+24}{r}$

The minimum cost occurs when the derivative of the Cost =0.

$C^{'}(r)=\frac{6\pi r^3-600}{25r^2}$

$6\pi r^3-600=0$

$6\pi r^3=600$

$\pi r^3=100$

$r^3=\frac{100}{\pi}$

$r^3=31.83$

r=3.17 cm

Recall that:

$h=\frac{400}{\pi r^2}$

$h=\frac{400}{\pi *3.17^2}$

h=12.67cm

The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.

3 0

3 years ago

Marvin scored 96, 93, 97, and 92 on four out of five math tests. In order to receive the highest grade in the class, he must obt

castortr0y [4]

Answer:98

Step-by-step explanation:

has to be higher then 97 so it has to be 98

5 0

2 years ago

Read 2 more answers

Math help mee plzzzzzz THERE IS A ATTACHMENT SO WAIT AT LEAST 30 SECONDS PLZZZZ

yarga [219]

I hope this helps you

4 0

2 years ago

Read 2 more answers

a) find center of mass of a solid of constant density bounded below by the paraboloid z=x^2+y^2 and above by the plane z=4.(b) F

slava [35]

Answer: x-bar = y-bar = 0 whereas z-bar = 8/3

M= (c^2)/8 which is intern equal to 2 $\sqrt{2}$

Step-by-step explanation:

Find the area, by setting the limits as

= $4\cdot \int _0^{\frac{\pi }{2}}\int _0^2\int _{r^2}^4\:rdzdrd\theta$

$=4\cdot \int _0^{\frac{\pi }{2}}\int _0^2r\cdot \:4-r^3drd\theta$

$=4\cdot \int _0^{\frac{\pi }{2}}4d\theta$

$=8\pi$

Therefore;

Mxy= $\int _0^{2\pi }\int _0^2\int _{r^2}^4\:zrdzdrd\theta$

z-bar = 8/3

M= 8 $\pi$ dividing it into two volume gives us = 4 $\pi$

means $4\pi =\int _0^{2\pi }\int _0^{\sqrt{c}}\int _r^c\:rdzdrd\theta$

$4\pi =\left(\pi c^2-2\pi \frac{c^{\frac{3}{2}}}{3}\right)$

c=2 $\sqrt{2}$

5 0

3 years ago