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2 years ago

15

Kams monthly budget includes 249 for food, 160 for gasoline, and 150 for utilities. If he earns 1506 per month after taxes, how

much money is left for other expenses ?

1 answer:

seropon [69]2 years ago

5 0

947 because when we add the amount of food, gasoline, and utilities we get 559 and then 1506-559=947$

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Use the Law of Sines to solve for x. Round to the nearest whole number.

bagirrra123 [75]

Answer:

12.94

Step-by-step explanation:

By the law of sines:

x / sin(35) = 7.9 / sin70

x = 7.9 * sin(70)/sin(35)

= 7.9 * 0.9397 / 0.5736

= 12.94

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2 years ago

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3 years ago

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an English teacher would like to divide 15 boys and 6 girls into groups each with the same combination of boys and girls and nob

tamaranim1 [39]

Answer:

7groups

Step-by-step explanation:

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2 years ago

Suppose that a cyclist began a 374 mi ride across a state at the western edge of the state, at the same time that a car traveli

kherson [118]

This is the concept of relative speed; We are required to calculate the speed of the car and the bicycle.
Distance between the car and Bicycle=374 miles
Time they met=5.5 hr
Speed traveled by bicycle=x
Speed traveled by car=x+33.4334
Relative speed=x+(x+33.4334)=(2x+33.4334) mph
Distance=speed*time
374=(2x+33.4334)*5.5
374=11x+183.8837
collecting like term we get:
374-183.8837=11x
11x=190.1163
thus;
x=(190.1163)/(11)
x=17.2833 mph
thus the speed of the bicycle was x=17.2833 mph
The speed of the car was (x+33.4334)=(17.2833+33.4334)=50.7167 mph

4 0

3 years ago

Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be

Travka [436]

Answer:

D = L/k

Step-by-step explanation:

Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is

dA/dt = in flow - out flow

Since litter falls at a constant rate of L grams per square meter per year, in flow = L

Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow

So,

dA/dt = in flow - out flow

dA/dt = L - Ak

Separating the variables, we have

dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

1/k∫-kdA/(L - Ak) = ∫dt

1/k㏑(L - Ak) = t + C

㏑(L - Ak) = kt + kC

㏑(L - Ak) = kt + C' (C' = kC)

taking exponents of both sides, we have

$L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt} (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k} - \frac{C"}{k} e^{kt}$

When t = 0, A(0) = 0 (since the forest floor is initially clear)

$A = \frac{L}{k} - \frac{C"}{k} e^{kt}\\0 = \frac{L}{k} - \frac{C"}{k} e^{k0}\\0 = \frac{L}{k} - \frac{C"}{k} e^{0}\\\frac{L}{k} = \frac{C"}{k} \\C" = L$

$A = \frac{L}{k} - \frac{L}{k} e^{kt}$

So, D = R - A =

$D = \frac{L}{k} - \frac{L}{k} - \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{kt}$

when t = 0(at initial time), the initial value of D =

$D = \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{k0}\\D = \frac{L}{k} e^{0}\\D = \frac{L}{k}$

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2 years ago