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Kamila [148]
2 years ago
11

The students in a third-grade classroom arranged the desks in this shape.

Mathematics
1 answer:
Eva8 [605]2 years ago
7 0

Answer:

D or B i think

Step-by-step explanation:

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Rotate angle ABC 90 degrees clockwise about the origin and then reflect over the x-axis
Lady_Fox [76]

Rotation and reflection are instances of transformation

See attachment for the new position of  \triangle ABC

From the question, we have:

A = (-1,1)

C = (-2,4)

The rule of 90 degrees clockwise rotation is:

(x,y) \to (y,-x)

Using the above transformation, the new points would be

A' = (1,1)

B' =(3,3)

C' = (4,2)

The next transformation is reflection over the x-axis

The rule of this transformation is:

(x,y) \to (x,-y)

So, the new points would be:

A" = (1,-1)

B" = (3,-3)

C" = (4,-2)

See attachment for the new points

Read more about transformation at:

brainly.com/question/11709244

8 0
2 years ago
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xxTIMURxx [149]

Answer:

I got 2 km per hour

Step-by-step explanation:

7 0
3 years ago
sue has 18 sweets, tony also has 18 sweets, sue gives tony x sweets, sue eats 5 of her sweets, tony then eats half of his sweets
Tomtit [17]

Answer:

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7 0
3 years ago
Read 2 more answers
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
A student is given a rock that is known to have a mass of 436.8 grams. She measures the mass of the rock three different times w
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I’m not sure just need to answer questions to help me so sorry
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