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3 years ago

Write 12 digit whole numbers with the following digits in correct places

Mathematics

1 answer:

lbvjy [14]3 years ago

4 0

Answer:

29,675,000

Step-by-step explanation:

i’m sorry if this is wrong but i think it’s right

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What is the surface area of the rectangular prism?

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3 years ago

Rich has a bag of 24 pencils. Each pencil is 19 centimeters long. What is the combined length of the pencils in meters

Roman55 [17]

Multiply the number of pencils by the length to get total centimeters first, then convert to meters:

24 x 19 = 456 cams.

1 meter = 100 cm

Divide total cm by 100 to get meters

456 / 100 = 4.56 meters

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3 years ago

Brahmagupta’s solution to a quadratic equation of the form ax2 + bx = c involved only one solution. Which solution would he have

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Answer:

Step-by-step explanation

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3 years ago

Read 2 more answers

When a breeding group of animals is introduced into a restricted area such as a wildlife reserve, the population can be expected

jasenka [17]

Answer:

A. Initially, there were 12 deer.

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. After 15 years, there will be 410 deer.

D. The deer population incresed by 30 specimens.

Step-by-step explanation:

$N=\frac{12.36}{0.03+0.55^t}$

The amount of deer that were initally in the reserve corresponds to the value of N when t=0

$N=\frac{12.36}{0.33+0.55^0}$

$N=\frac{12.36}{0.03+1} =\frac{12.36}{1.03} = 12$

A. Initially, there were 12 deer.

B. $N(10)=\frac{12.36}{0.03 + 0.55^t} =\frac{12.36}{0.03 + 0.0025}=\frac{12.36}{y}=380$

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. $N(15)=\frac{12.36}{0.03+0.55^15}=\frac{12.36}{0.03 + 0.00013}=\frac{12.36}{0.03013}= 410$

C. After 15 years, there will be 410 deer.

D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:

ΔN=N(15)-N(10)

ΔN=410 deer - 380 deer

ΔN= 30 deer.

D. The deer population incresed by 30 specimens.

8 0

3 years ago

A sample of radium has a weight of 1.5 mg and decays

Oliga [24]

Answer: $(b)$

Step-by-step explanation:

Given

Initial mass of radium is $A_o=1.5\ mg$

It decays by half every 6 years i.e. $T_{\frac{1}{2}}=6$

For radioactive decay, sample at any time is given by

$\Rightarrow A=A_o2^{-\dfrac{t}{T_{\frac{1}{2}}}}$

Insert the values

$\Rightarrow A=1.5\cdot 2^{\dfrac{-t}{6}}\\\\\Rightarrow A=1.5\cdot \left(\dfrac{1}{2}\right)^{\dfrac{-t}{6}}\\\\\Rightarrow A=1.5\cdot (0.5)^{\dfrac{t}{6}}$

Thus, option (b) is correct.

6 0

3 years ago