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2 years ago

1. Match each diagram with its name.

Mathematics

1 answer:

aev [14]2 years ago

4 0

. The entire answers are infront of you.

Below are the answers. Ontop are the questions.

You just need to see what letters and arrow matches the answer in the question.

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a garden is in a shape of a rectangle ABCD and a semi circle AD the diameter of the semi circle carol is going to cover the gard

ser-zykov [4K]

The shape of the garden is composite figure of a rectangle and a semicircle. The area of that shape will be
Area = AB x CD + π x AD² / 4
Once the area is calculated, the number of boxes of fertilizer is determined
Area / 12 m = Number of boxes
After getting the number of boxes, the total cost can be determined
£4.99 x Number of boxes

5 0

2 years ago

Which expression is equivalent to is m4? m2+m3 m1+m2 m1+m3 m5+ m6

Airida [17]

Answer: m1+m3

Step-by-step explanation: 1+3=4 and m1+m3 has the same varibles in the expression.

8 0

2 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

2 years ago

What’s the inequality of x-3>8

sergiy2304 [10]

Answer:

x>11

Step-by-step explanation:

To simplify this inequality, we add 3 from both sides to get x-3+3>8+3. Since -3+3 cancels out to 0, we are left with x>8+3. 8+3=11 so x>11.

Hope this helps! If you need further explanation let me know :)

3 0

1 year ago

the first group of students consists of 10 and their average age was 13 years old the next group consisted of 30 students and th

Kazeer [188]

Answer: The average was 9 years old find the average age of both groups is 10 years old.

Step-by-step explanation:

Formula foe average : $\text{Average} =\dfrac{\text{Sum of all observations}}{\text{Number of observations}}$

Given : The first group of students consists of 10 and their average age was 13 years old.

i.e. $13=\dfrac{\text{Sum of ages of students of first group}}{10}\\\\\Rightarrow\ \text{Sum of ages of students of first group}=13\times10=130$ (1)

The next group consisted of 30 students and their average was 9 years old.

i.e. $9=\dfrac{\text{Sum of ages of students of next group}}{30}\\\\\Rightarrow\ \text{Sum of ages of students of nextgroup}=9\times30=270$ (2)

Then from (1) and (2) , the sum of both groups (first group and next group )students = 130+270 =400

Combined students of both groups (first and next group )= 10+30=40

Now , the average of both groups = $\dfrac{\text{Sum of combined students}}{\text{combined students}}$

$=\dfrac{400}{40}=10$

Hence, the average was 9 years old find the average age of both groups is 10 years old.

4 0

3 years ago