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olga2289 [7]
3 years ago
6

Is the simplified form of 2 square root of 3 − 2 square root of 3 rational? (1 point)

Mathematics
2 answers:
uranmaximum [27]3 years ago
7 0
Yes the answer is rational.
tia_tia [17]3 years ago
7 0

Answer:

<h2>Yes.</h2>

Step-by-step explanation:

The given expression is: 2\sqrt{3}-2\sqrt{3}

Its simplified expression is zero, because they are equal but opposite. So, we know that zero is included inside the Rational Numbers Field, so the answer is yes, the simplified expression is rational.

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I lost my dad in the war
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If the lowest temperature in Whitesboro for the week was -1°F and the highest temperature was 20°F, what was the total change in
Talja [164]
Answer: The total change is 21.

Since you want to find the change from the lowest to highest, make it into a subtraction problem. First of all, take the smallest number in front of the largest.

20 -1

Next, add a minus sign in between, your equation should look like this:

20 - (-1) = x

Then, you cancel out the negative signs. After that you solve it

20 - (-1) -> 20 + 1.

Lastly, you add them together.

20 + 1 = 21.

In conclusion, the answer is 21.

3 0
3 years ago
Read 2 more answers
24+82 rounding or compatible numbers to estimate the sum
allsm [11]

Answer:

106

Step-by-step explanation:

you take 82 and 24 and add them together so it would be like this 2 + 4 = 6 and 8 + 2 = 10 so the 8 is in a tens place so the answer from adding 8 + 2 = 100 because you in the tens place so the final answer would be 106 after adding it all up.

4 0
2 years ago
PLEASE HELP ASAP BY 11/12 BY 9:00 am!!!! WILL AWARD BRAINLIEST!!!
Eduardwww [97]

Answer:

x=1/4 y=1/4

Step-by-step explanation:

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8 0
3 years ago
At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is
Hitman42 [59]

Answer:

answer = 12.87 km/h

Step-by-step explanation:

Given

Ship A is sailing east at 25 km/h = \frac{dx}{dt}

ship B is sailing north at 20 km/h =\frac{dy}{dt}

here x and y are the  sailing at t = 4 : 00 pm for ship A and B respectively

so we get x = 4 ×25 =100 km/h

                 y = 4× 20 = 80 km/h

let z is the distance between the ships, we need to find \frac{dz}{dt} at t = 4 hr

At noon, ship A is 130 km west of ship B (12:00 pm)

so equation will be

z^2 = (130-x)^2 + y^2......................(i)\\put x = 100 and y = 80 \\\\we |  | get \\

z^2 = 30^2 + 80^2\\z =\sqrt{7300} km/h

derivative first equation w . r. to t we get

2z\frac{dz}{dt} =-2(130-x)\frac{dx}{dt}+2y\frac{dy}{dt}

\frac{dz}{dt} =\frac{1}{z}[(x -130)\frac{dx}{dt} +y\frac{dy}{dt}]

\frac{dz}{dt} = \frac{( -20\times25 + 80\times20)}{\sqrt{7300} }

     = \frac{1100}{85.44}\\  = 12.87km/h

8 0
3 years ago
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