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3 years ago

Turtles tutles turtles turtles

Mathematics

2 answers:

Paul [167]3 years ago

7 0

Answer:

hey

Step-by-step explanation:

I like ya cut g XD

lesantik [10]3 years ago

4 0

Answer:

TURTLES TURTLES bro!!!!!!!

Step-by-step explanation:

my pet turtles name is godzilla. that deserves the crown i think.

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Please help with this

Dafna1 [17]

Answer:

Step-by-step explanation:

● first one:

The diagonals of a rhombus are perpendicular to each others wich means that they form four right angles.

STP is one of them so this statement is true.

● second one:

If ST and PT were equal this would be a square not a rhombus.

● third one:

If SPQ was a right angle, this woukd be a square.

● fourth one:

Again if the diagonals SQ and PR were equal, this would be a square.

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3 years ago

Will you do my math homwork

gayaneshka [121]

If you post your questions we could help

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3 years ago

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Look at triangles TUV and T'U'V'.Which best explains the transformation of triangle TUV to form triangle T'U'V'?

Iteru [2.4K]

Answer:

The answer is D or Triangle TUV is dilated by a scale factor of 9 and then translated 2 units right and 1 unit down to form triangle T"U"V" . I just took the test and I hope I helped : )

Step-by-step explanation:

6 0

3 years ago

HOW MANNY YARDS ARE IN A FEET

Vladimir [108]

Answer:

0.333333 yards in a foot

Step-by-step explanation:

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3 years ago

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Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

5 0

3 years ago