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2 years ago

Are you guys all in k12? I’m an 8th grader

Mathematics

2 answers:

antoniya [11.8K]2 years ago

6 0

I am in 6th grade honors

masya89 [10]2 years ago

6 0

Im a 9th grader. trust me high school is so much fun.

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What are the dimensions of the simple shapes you obtained in part C? List dimensions for all your decompositions (sets of shapes

olasank [31]

Answer:

Option 1

Figure Length (feet) Width (feet)

small rectangle 14 6

large rectangle 20 7

Figure Base (feet) Height (feet)

triangle 6 6

Option 2

Figure Length (feet) Width (feet)

small rectangle 6 7

large rectangle 14 13

Figure Base (feet) Height (feet)

triangle 6 6

Step-by-step explanation:

3 0

2 years ago

The distance from Earth to Mars is 136,000,000 miles. A spaceship travels at 31,000 miles per

posledela

136,000,000/31,000= how many hours for the spaceship to reach mars
136,000,000/31,000/24= how many days for the spaceship to reach mars
136,000,000/31,000/24 is approx. 182.7 days or 183 days (rounded up b/c of "to the nearest day").

7 0

2 years ago

Read 2 more answers

Kevin made a jewelry box with dimensions shown 5cm,10cm,15cm what is the value of the jewelry box?

zaharov [31]

Answer:

750

Step-by-step explanation:

10 times 5 times 15

3 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

Order the numbers from least to greatest. 8, 0, -8, -3, 5

Kitty [74]

Answer:

-8,-3,0,5,8

Step-by-step explanation:

Number the the negatives from greatest to least first and continue to number the positives from least to greatest.

7 0

3 years ago