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3 years ago

Please help me with this

Mathematics

1 answer:

AleksAgata [21]3 years ago

4 0

Count the number of digits after the first digit. this gives you the exponent on the 10. The first number in scientific form has the decimal point after the first digit.

For mercury there are 7 digits after the 5:-

So the Mercury answer is

5.7909175 * 10^7

Follow this procedure for the other planets

The answer for Venus is

1.08208930 * 10^8

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You are recreating a scale drawing at a different scale. The original scale is 1 cm : 8 m. The new scale is 1 cm : 2 m. In the o

densk [106]

Answer:

The segment should be 36 centimeters long

Step-by-step explanation:

Let us use the ratio method to solve the question

∵ The original scale is 1 cm : 8 m

∵ In the original drawing, the length of a segment is 9 centimeters

→ By using the ratio method

→ cm : m

→ 1 : 8

→ 9 : x

→ By using the cross multiplication

∵ 1 × x = 9 × 8

∴ x = 72 m

∴ The actual length of the segment is 72 m

∵ The new scale is 1 cm : 2 m

∵ The actual length of the segment is 72 m

→ By using the ratio method

→ cm : m

→ 1 : 2

→ y : 72

→ By using the cross multiplication

∵ y × 2 = 1 × 72

∴ 2y = 72

→ Divide both sides by 2 to find y

∴ y = 36

∴ The drawing length of the segment is 36 cm in the new scale

∴ The segment should be 36 centimeters long

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3 years ago

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Step-by-step explanation:

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2 years ago

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3 years ago

Read 2 more answers

Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$

Which has the following solution:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that $Q(6) = (1-0.03)Q(0) = 0.97Q(0)$ . We use this to find r.

$Q(t) = Q(0)e^{-rt}$

$0.97Q(0) = Q(0)e^{-6r}$

$e^{-6r} = 0.97$

$\ln{e^{-6r}} = \ln{0.97}$

$-6r = \ln{0.97}$

$r = -\frac{\ln{0.97}}{6}$

$r = 0.0051$

$Q(t) = Q(0)e^{-0.0051t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0051t}$

$0.5Q(0) = Q(0)e^{-0.0051t}$

$e^{-0.0051t} = 0.5$

$\ln{e^{-0.0051t}} = \ln{0.5}$

$-0.0051t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.0051}$

$t = 135.9$

The half-life of the radioactive substance is 135.9 hours.

6 0

2 years ago