I need help pls. Question 11

Brandon earns a rebate each month on his meals and movie tickets. He earns $0.50 for each movie ticket and $2 for each meal. Las

Alex787 [66]

A. 25 = (.5 * 22) + (2 * x)

B. 25 = (.5 * 22) + (2 * x)
25 = 11 + 2x
14 = 2x
7 = x

C. 7

7 0

3 years ago

A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th

Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let $S(t)$ be the amount of sugar in the tank at time $t$ . Then $S(0)=25$ .

Sugar is added to the tank at a rate of P lb/min, and removed at a rate of

$\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}$

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

$\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}$

Separate variables, integrate, and solve for S.

$\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt$

$\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt$

$-100\ln\left|P-\dfrac S{100}\right|=t+C$

$\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t$

$P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}$

$\dfrac S{100}=P-Ce^{-100t}$

$S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}$

Use the initial value to solve for C :

$S(0)=25\implies 25=100P-C\implies C=100P-25$

$\implies S(t)=100P-(100P-25)e^{-100t}$

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

$1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}$

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick P such that

$S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}$

5 0

2 years ago

Gena wants to estimate the quotient of –21.87 divided by 4.79. Which expression shows the estimate using front-end estimation?

allochka39001 [22]

-21/5 is the closest!!!

Hope this helps.

If you want me to explain why, just ask! :)

7 0

2 years ago

Read 2 more answers

AB≅BC≅CD≅DE, AD = 8x - 4, AB = x + 2 solve for AB

Levart [38]

Just solve for , then substitute back into either expression and calculate either BC or AD. The other one is then the same amount.

The one about the square is the same thing except you don't care how the figure is named because all 4 sides are equal anyway. Just set the two expressions equal to one another and then solve for

4 0

3 years ago

Read 2 more answers

Triangle A B C is shown. Angle A C B is a right angle. The length of hypotenuse A B is 12 centimeters, the length of C B is 9.8

RSB [31]

Answer: We can find angle BAC by using (1) SinA = 9.8/12

(2) CosA = 6.9/12

(3) TanA = 9.8/6.9

Step-by-step explanation: The question indicates that we have a right angled triangle ABC with the right angle at point C (that is, angle ACB is the right angle). Also the three sides have been labeled as AB equals 12 cm, CB equals 9.8 cm and AC equals 6.9 cm. Line AB has been identified as the hypotenuse and angle A (that is, BAC) is the reference angle. If angle A is the reference angle, then line CB facing angle A shall be the opposite side, while line AC is the adjacent (which lies between the reference angle and the right angle). Please see the attached diagram for details.

Having these details available, we can actually find angle A by using any of the three trigonometric ratios, since all three sides are given. Hence,

SinA = opposite/hypotenuse

SinA = 9.8/12

SinA = 0.8167

Checking with your calculator or table of values, A = 54.76 (approximately 55)

Also CosA = adjacent/hypotenuse

CosA = 6.9/12

CosA = 0.5750

Checking with your calculator or table of values A = 54.90 (approximately 55).

Finally TanA = opposite/adjacent

TanA = 9.8/6.9

TanA = 1.4203

Checking with your calculator or table of values A = 54.83 (approximately 55).

3 0

3 years ago