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LiRa [457]
3 years ago
15

Help me with right answer plz

Mathematics
1 answer:
givi [52]3 years ago
7 0

Answer:

13.5

Step-by-step explanation:

area: 1/2*[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]

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Need help on 5 and 6
lianna [129]
Well the equation for slope is \frac{ y_{2}- y_{1}  }{ x_{2} - x_{1}  }

number 5
\frac{-1-(-2)}{3-(-2)} = \frac{1}{5}
which means the slope is \frac{1}{5}
5 0
3 years ago
I need help with all of these
Olin [163]
If you take your calculator and enter in -82+28 it should give you the answer easily If you dont have one use ur phone.

Easiest way 1.o.1
6 0
4 years ago
Solve the equation. 8n+15=25
Aneli [31]
N=1.25 or you can use n= 1 1/4
3 0
3 years ago
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Help with numer 5 please. thank you​
Alex17521 [72]

Answer:

See Below.

Step-by-step explanation:

We are given that:

\displaystyle I = I_0 e^{-kt}

Where <em>I₀</em> and <em>k</em> are constants.

And we want to prove that:

\displaystyle \frac{dI}{dt}+kI=0

From the original equation, take the derivative of both sides with respect to <em>t</em>. Hence:

\displaystyle \frac{d}{dt}\left[I\right] = \frac{d}{dt}\left[I_0e^{-kt}\right]

Differentiate. Since <em>I₀ </em>is a constant:

\displaystyle \frac{dI}{dt} = I_0\left(\frac{d}{dt}\left[ e^{-kt}\right]\right)

Using the chain rule:

\displaystyle \frac{dI}{dt} = I_0\left(-ke^{-kt}\right)  = -kI_0e^{-kt}

We have:

\displaystyle \frac{dI}{dt}+kI=0

Substitute:

\displaystyle \left(-kI_0e^{-kt}\right) + k\left(I_0e^{-kt}\right) = 0

Distribute and simplify:

\displaystyle -kI_0e^{-kt} + kI_0e^{-kt} = 0 \stackrel{\checkmark}{=}0

Hence proven.

4 0
3 years ago
Question is in photo.<br> Thanks
Triss [41]

Answer: yes

Step-by-step explanation: because if you can draw a line through the graph and hit most of the dots it would be considered linear.

8 0
3 years ago
Read 2 more answers
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