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yanalaym [24]
2 years ago
13

HelpActivity:2x²2+2x-312=0​

Mathematics
1 answer:
SIZIF [17.4K]2 years ago
7 0

Answer:

Step-by-step explanation:

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The last graph provided is the correct one. With points (-1,-3) (-3,0) (-3,-3)
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Margarita [4]

Answer:

Answers A (1/3), C (4/15) and D (5/6) are all repeating decimals.  

Step-by-step explanation:

Repeating decimals are decimals in which a number or sequence of numbers is repeated over and over when dividing the numerator by the denominator of a fraction.  For example, when you divide 1 by 3 (1/3), you will get a repeating decimal of 0.3333...since 3 goes into 10 three times with a remainder of 1 and will keep going.  Likewise, when you divide 4/15, you get an initial value of 0.2, with a repeating 6, or 0.266666...  Lastly, when you divide 5/6, you will get a repeating decimal of 0.833333....  The other answers will all be terminating decimals when you divide the numerator by the denominator.  Terminating decimals mean the stop at some point and don't continue.  

6 0
3 years ago
Read 2 more answers
13 If CDEF is a rhombus, find m FED.
Tatiana [17]

Answer:

36

Step-by-step explanation:

8x-20=5x+1

-5       -5

3x=-20+1

+20 +20

3x=21

3  3

x=7

8(7)-20

FED=36

8 0
2 years ago
A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t)=(t)ln(5t). Find the acceleration of the par
lyudmila [28]

Answer:

a(\frac{1}{5e})=5e

Step-by-step explanation:

we are given equation for position function as

s(t)=tln(5t)

Since, we have to find acceleration

For finding acceleration , we will find second derivative

s'(t)=\frac{d}{dt}\left(t\ln \left(5t\right)\right)

=\frac{d}{dt}\left(t\right)\ln \left(5t\right)+\frac{d}{dt}\left(\ln \left(5t\right)\right)t

=1\cdot \ln \left(5t\right)+\frac{1}{t}t

s'(t)=\ln \left(5t\right)+1

now, we can find derivative again

s''(t)=\frac{d}{dt}\left(\ln \left(5t\right)+1\right)

=\frac{d}{dt}\left(\ln \left(5t\right)\right)+\frac{d}{dt}\left(1\right)

=\frac{1}{t}+0

a(t)=\frac{1}{t}

Firstly, we will set velocity =0

and then we can solve for t

v(t)=s'(t)=\ln \left(5t\right)+1=0

we get

t=\frac{1}{5e}

now, we can plug that into acceleration

and we get

a(\frac{1}{5e})=\frac{1}{\frac{1}{5e}}

a(\frac{1}{5e})=5e


5 0
3 years ago
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