Find the dependent and independent variable in r = 5s + 3.

Yves bought 420 tropical fish for a museum display. He bought 6 times as many parrotfish as angelfish. How many of each type of

ohaa [14]

The answer would be d , because you have to add how many of each fish to equal up to the sum of 420 . which is x y . so the first equation is x + y = 420 . then it says he bought 6 times as many parrotfish as he did angelfish so you would have to multiply x which represents angelfish by 6 . to get the second equation of y = 6x .

5 0

1 year ago

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use distributive property to simplify the expression 3 (8x + 2y) I need help with my algebra homework someone please help.

inysia [295]

$6 \times (4 \times x + y)$

4 0

1 year ago

You are traveling down a country road at a rate of 95 feet/sec when you see a large cow 300 feet in front of you and directly in

emmainna [20.7K]

Answer:

1) You can rely solely on your brakes because when doing so the car will just travel 250ft from the point you hit your brakes till the point the car stopped completely, leaving you 50ft away from the cow.

2) See attached picture.

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in $ft/s^{2}$

3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) $j(t)=\left \{ {{95t-9t^{2}; 0\le t$

$j'(t)=\left \{ {{95-18t; 0\leq t$

(see attached picture for graph)

Step-by-step explanation:

1) In this part of the problem we need to find the time when the speed of the car is 0. Gets to a complete stop. For this we will need to take the derivative of the position function so we get:

$j(t)=95t-9t^2$

$j'(t)=95-18t$

and we set the first derivative equal to zero so we get:

95-18t=0

and solve for t

-18t=-95

$t=\frac{95}{18}$

t=5.28s

so now we calculate the position of the car after 5.28 seconds, so we get:

$j(5.28)=95(5.28)-9(5.28)^{2}$

$j(5.28)=250.69ft$

so we have that the car will stop 250.69ft after he hit the brakes, so there will be about 50ft between the car and the cow when the car stops completely, so he can rely just on the breaks.

2) For answer 2 I take the second derivative of the function so I get:

$j(t)=95t-9t^{2}$