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djyliett [7]
2 years ago
12

Which of the functions listed below will produce the graph shown here? HELP!!!!

Mathematics
1 answer:
aivan3 [116]2 years ago
8 0
A) -3x^2/2x-1 is the right answer
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What percent of $375 is $11.25 and what percent of $56.25 is $11.25?
Angelina_Jolie [31]

Answer:

3%percent of $375 is $11.25 and 20% of $56.25 is $11.25 .

Step-by-step explanation:

Formula

Percentage = \frac{Part\ value\times 100}{Total\ values}

First Part

As given

Total value = $375

Part value = $ 11.25

Put values in the formula

Percentage = \frac{11.25\times 100}{375}

Percentage = \frac{1125}{375}

Percentage = 3 %

Second Part

Total value = $56.25

Part value = $11.25

Put the values in the formula

Percentage = \frac{11.25\times 100}{56.25}

Percentage = \frac{1125}{56.25}

Percentage = 20%

Therefore 3%percent of $375 is $11.25 and 20% of $56.25 is $11.25 .

3 0
3 years ago
Read 2 more answers
Type the correct question of the line
Pepsi [2]
The correct equation of the line is Y=-2x-1 :)
7 0
1 year ago
The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s
Norma-Jean [14]

The positions when the particle reverses direction are:

s(t_1)=55ft\\\\s(t_2)=28ft

The acceleraton of the paticle when reverses direction is:

a(t_1)=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=18\frac{ft}{s^{2}}

Why?

To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

Also, when the particle reverses its direction, the velocity is equal to 0.

We are given the following function:

s(t)=2t^{3}-21t^{2}+60t+3

So,

- Derivating to get the velocity function, we have:

v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60

Now, making the function equal to 0, to find the times when the particle reversed its direction, we have:

v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s

We know that the particle reversed its direction two times.

- Derivating the velocity function to find the acceleration function, we have:

a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42

Now, substituting the times to calculate the accelerations, we have:

a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}

Now, substitutitng the times to calculate the positions, we have:

s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft

Have a nice day!

3 0
3 years ago
Which statement is false? A. (−2)×5<(−20) B. (−2)×(−5)>(−25) C. 2×5>(−25) D. 2×(−5)<20
natali 33 [55]

Answer:

(−2)×5<(−20)

Step-by-step explanation:

Evaluating the options given :

(−2)×5<(−20)

Open the bracket

- 10 < - 20 (false) ` This expression isn't true

(−2)×(−5)>(−25)

Open the bracket

10 > - 25 (true)

2×5>(−25)

Open the bracket

10 > - 25 (true)

2×(−5)<20

Open the bracket

- 10 < 20

4 0
2 years ago
Is -7-5+3=1 true or false
grigory [225]
-7-5+3=1
-12+3=1
-9≠1. As a result, this is False. Hope it help!
7 0
2 years ago
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