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RideAnS [48]
1 year ago
7

the function f(x)=2•5x can be used to represent the curve through the points (1,10) (2,50) and (3,250). What is the multiplicati

ve rate of change of the function.
Mathematics
2 answers:
MaRussiya [10]1 year ago
11 0

Answer:- the multiplicative rate of change of the function=5


Explanation:-

Given function: f(x)=2\cdot5^x on coparing it with exponential function Ab^x,here A=2 is the initial amount of the function and b=5 is the multiplicative rate of growth.

Alternatively,

Given points=(1,10) (2,50) and (3,250)

We can see that

at x=1, y=10=2×5

at x=2, y=50=2×5×5

at x=3, y=250=2×5×5×5

⇒  the multiplicative rate of change of the function=5

Annette [7]1 year ago
8 0
The answer would be 5
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pickupchik [31]
My answer is: D. <span>(6,0,0)

Given: 
</span><span> 7x +2y +3z =42

I assumed that the format in the given choices is (x,y,z). So, I substituted each number to its corresponding variable.

A. </span>(14,0,0)  → 7(14) +2(0) +3(0) = 42 → 98 + 0 + 0 ≠ 42  NOT THE ANSWER<span>
B. (7,0,0) </span>→ 7(7) +2(0) +3(0) = 42 → 49 + 0 + 0 ≠ 42 NOT THE ANSWER<span>
C. (21,0,0) </span>→ 7(21) +2(0) +3(0) = 42 → 147 + 0 + 0 ≠ 42 NOT THE ANSWER<span>
D. (6,0,0) </span>→ 7(6) +2(0) +3(0) = 42 → 42 + 0 + 0 = 42 CORRECT ANSWER.


<span>The ordered triples indicated where the plane cuts the x-axis for this equation is D. (6,0,0). </span>
4 0
2 years ago
Read 2 more answers
A sumo wrestling ring is circular and has a circumference of 4.6\pi \text{ meters}4.6π meters4, point, 6, pi, start text, space,
Zina [86]

Answer:

The area of the wrestling ring is 5.29π m^2

Step-by-step explanation:

A sumo wrestling ring is circular and has a circumference of 4.6π meters.

To find its area, we need to first find its radius.

The circumference of a circle is given as:

C = 2 \pi r

where r = radius

Hence, the radius of the ring will be:

4.6\pi = 2\pi r\\r = 4.6 /2 = 2.3m

The radius of the ring is 2.3 m.

The area of a circle is given as:

A = \pi r^2

Hence, the area of the wrestling ring (in terms of π) will be:

A = \pi * 2.3^2

A = 5.29\pi m^2

The area of the wrestling ring is 5.29π m^2

7 0
2 years ago
A while back, either James borrowed $12 from his friend Rita or she borrowed $12 from him, but he can’t quite remember which. Ei
Tamiku [17]

Answer:

First option is correct.

Step-by-step explanation:

Let amounts of money James may have after he sees Rita be x.

It is given that either James borrowed $12 from his friend Rita or she borrowed $12 from him.

So meeting Rita second time either he will $12 or he will receive $12.

The amounts of money James may have after second meeting with Rita is:

Case 1: If he pays $12.

x=42.80-12             .... (1)

Case 1: If he receives $12.

x=42.80+12           .... (2)

On combining equation (1) and (2), we get

x=42.80\pm 12

Subtract 42.80 from both sides.

x-42.80=\pm 12

It can be written as

|x-42.80|=12

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4 0
1 year ago
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Answer:

C, they are the same

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7 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}

j'(t)=95-18t

j"(t)=-18

and then we graph them. (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)

5 0
2 years ago
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