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xenn [34]
4 years ago
14

Which pairs of angles in the figure below are vertical angles?

Mathematics
2 answers:
SIZIF [17.4K]4 years ago
7 0

Answer:

B and C

Step-by-step explanation:

Vertical angles are the angles opposite of each other.

amm18124 years ago
5 0

Answer: ISN and TSW or TSN and ISW same thing so b and c

You might be interested in
Michael is 5 years old than rui feng and. Vishal is thrice as old as micheal if rui feng is 3 years old find Michael present age
lorasvet [3.4K]

Answer:

a) Micheal's present age = 8 years old

b) The sum of their ages in 6 years time = 63 years.

Step-by-step explanation:

From the above question, we know that

Rui feng is 3 years old

We are told that

• Michael is 5 years old than Rui feng

Micheal's present age is calculated as:

= 5 + Rui feng's age

= 5 + 3 = 8 years

• Vishal is thrice as old as Micheal

Vishal's present age is calculated as

= 3(Micheal age)

= 3(8)

= 24 years.

The sum of their ages in six years time

= ( 3 + 6) +( 8 + 6) + ( 24 + 6)

= 18 + 15 + 30

= 63 years

Therefore, a) Micheal's present age = 8 years old

b) The sum of their ages in 6 years time = 63 years.

3 0
3 years ago
7 - 5n = 6 + 6(n + 2)
postnew [5]
7- 5n=6+6(n+2)
7- 5n=6+6n+12
7 -5n=18+6n
5n-6n=18-7
-11n=11
n=-1
4 0
3 years ago
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
Can someone answer number 16 PLEASE
-Dominant- [34]

Answer:

If i am correct it should be 34

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Please answer this correctly
ArbitrLikvidat [17]

Answer:

326

Step-by-step explanation:

l x w

7x8

25x6

4x30

326

3 0
3 years ago
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