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vazorg [7]
2 years ago
10

Which of the following statements are Ture...?

Mathematics
1 answer:
Brilliant_brown [7]2 years ago
4 0

Answer:

1, 2, 4, 6 are true statements

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If x is a number that satisfies 4x+3=18, can x be equal to 3? explain.
Akimi4 [234]
The answer is no. because is you substitute the value 3 in for x into the equation and solve you get,

4(3)+3=18 which says 12+3=18 15=18 and that is an untrue statement. 
Hope this helps explain the concept so you understand! Any questions feel free to ask! Please rate if I helped you! thank you so much!!
4 0
2 years ago
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Elana runs for 28 seconds and finishes at 250 meters .what is her velocity
Sophie [7]

Answer:

8.9m/s

Step-by-step explanation:

Time= 28s

Displacement= 250m

Velocity=?

Velocity (v) = displacement (d)/ Time (t)

V= 250/28

V=8.9m/s

OR YOU CAN APPROXIMATE IT

V=8.928

YOU CAN APPROXIMATE IT TO

V=8.93m/s

3 0
3 years ago
Evaluate the equation. 193b = 212.3
trasher [3.6K]

Answer:

b= 1.1

Step-by-step explanation:

193b= 212.3

divide both sides by 193 to isolate b

212.3÷193=1.1

3 0
2 years ago
Writing g for the acceleration due to gravity, the period,T, of a pendulum of length l is given by
ivanzaharov [21]

Step-by-step explanation:

T=2\pi\sqrt{\frac{l}{g}}

T+\Delta T=2\pi\sqrt{\frac{(l+\Delta l)}{g}} =2\pi\sqrt{\frac{l}{g}}\sqrt{1+\frac{\Delta l}{l}}=2\pi\sqrt{\frac{l}{g}}(1+\frac{1}{2}\frac{\Delta l}{l}+0(\frac{\Delta l}{l}))=T(1+\frac{1}{2}\frac{\Delta l}{l}+0(\frac{\Delta l}{l}))

Here, the Taylor approximation for a square root was applied, and O(x) stands for all negligible terms of Taylor's sum with respect to variable x.

So, \Delta T=T\frac{1}{2}\frac{\Delta l}{l}

b. For an increase of 2%, that is:

\frac{\Delta l}{l}=0.02

\frac{\Delta T}{T}=\frac{1}{2}0.02=0.01=1\%

7 0
3 years ago
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<img src="https://tex.z-dn.net/?f=%5C%5C%20jn%20j%20j%20j" id="TexFormula1" title="\\ jn j j j" alt="\\ jn j j j" align="absmidd
labwork [276]

Answer:

glhIblzikdjbgjdzgbobrgbzoerbgoernbs

Step-by-step explanation:

ur welcome ;)

6 0
3 years ago
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