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sergejj [24]
4 years ago
14

If ln(x)=−0.123, what does x equal? Express your answer numerically using three significant figures.

Mathematics
1 answer:
Mrac [35]4 years ago
3 0

Answer:

these but jk I eont know girl or boy you asking the wiring person

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4. Lucy is k years old. She has a sister who is three years younger than she is and
WITCHER [35]

Answer:

4k - 8

Step-by-step explanation:

Lucy is k years old so we know k is an important part of this problem.
Sister 1 is 3 years younger, so she is k - 3 years old.
Sister 2 is 5 years less than twice Lucy's age therefore, she is 2k (twice Lucy's age) - 5 years old.
Now we have: k + (k - 3) + (2k - 5)
We can simplify like terms with the k's and get 4k. then we add (-3) and (-5) to get 4k - 8.

Hope this helps!

7 0
3 years 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
Jenny read 150 pages of a book in 5 hours. Which rate is proportional to the rate at which Jenny read?
erik [133]

Answer:

she would have read 30 pages per hour

Step-by-step explanation:

hope this helps

5 0
3 years ago
Put a positive factor back into the square root:
daser333 [38]

Answer:

-\sqrt{15}

Step-by-step explanation:

To put factor back into the square root, we have to put squared value.

like 6 can be written as \sqrt{36} .

Similarly, 5 can be written as \sqrt{25}

= -5\sqrt{0.6}

= -\sqrt{25*0.6}

= -\sqrt{15}

4 0
3 years ago
2/5 into a whole number
Advocard [28]

Answer:

2/5 its a quotient just 2 divide by 5 equals to 0.4

6 0
3 years ago
Read 2 more answers
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