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jeka94
2 years ago
13

Find the coordinates of the midpoint of1. (6,3) and (4,5)​

Mathematics
1 answer:
expeople1 [14]2 years ago
8 0

Answer:

(5,4)

Step-by-step explanation:

midpoint = (6 + 4 / 2 , 3 +5 / 2)

= (5,4)

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I have like 20 minutes please help<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D-7x%2B10%3D0" id="TexFormula1" title="x^{2}-7x+10
Mandarinka [93]

Answer: Hmm Ok i do this-

Step-by-step explanation:

Wait are we trying to find x?

Ok so for the first one x = 5 and for the second one x = 2

3 0
3 years ago
(w-9)8 need help please
nikitadnepr [17]
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8 0
3 years ago
Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volt
ser-zykov [4K]

Answer:

I(t)=\frac{1}{3}(1-e^{-30t})

Step-by-step explanation:

We are given that

\frac{dI}{dt}+\frac{R}{L}I=\frac{V(t)}{L}

R=150 ohm

L=5 H

V(t)=10 V

P=\frac{R}{L}=\frac{150}{5}=30

I.F=e^{\int Pdt}=e^{\int 30 dt}=e^{30 t}

I(t)\times I.F=\int e^{30 t}\times 10 dt+C

I(t)\times e^{30 t}=\frac{10}{30}e^{30 t}+C

I(t)=\frac{1}{3}+Ce^{-30 t}

I(0)=0

Substitute t=0

0=\frac{1}{3}+C

C=-\frac{1}{3}

Substitute the values

I(t)=\frac{1}{3}-\frac{1}{3}e^{-30 t}

I(t)=\frac{1}{3}(1-e^{-30t})

5 0
3 years ago
What are the solutions to the system <br> y=x^2-6x+7<br> y=-x+13
dolphi86 [110]
I've attached the graphs to this answer. I hope they help. 

3 0
3 years ago
Read 2 more answers
How does the graph of y=-3√2x-4 compare to the graph of y=-3√x-4?
kifflom [539]
<span>Assuming the graph is y=-3(√2x)-4 and y=-3√(x-4) the transformation would be:

</span><span>The graph is compressed horizontally by a factor of 2
x=1/2x'
</span>y=-3(√2x)-4 
y=-3(√x')-4 <span>

</span><span>moved left 4
x=x'-4
</span>y=-3(√x)-4 
y=-3(√x'-4)-4 
<span>
moved down 4
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</span>y=-3(√x-4)-4 
y'-4=-3(√x'-4)-4 
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y'=-3(√x'-4)

Answer: C. <span>The graph is compressed horizontally by a factor of 2, moved left 4, and moved down 4.
</span>
6 0
4 years ago
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