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3 years ago

Question 1

Mathematics

1 answer:

babunello [35]3 years ago

6 0

Answer:

The answer is B

I hope you got it right!!!

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Find the value of x that will make AB А 9x - 40 В. 3x + 20 x = [?]

Charra [1.4K]

Answer:

x = 10

Step-by-step explanation:

∵ Both angle ∠A and are corresponding angles.

∠A = ∠B

9x - 40 = 3x + 20

9x = 3x + 20 + 40

9x - 3x = 20 + 40

6x = 60

x = 60/6

x = 10

∴ x = [10]

7 0

3 years ago

In a research laboratory, a scientist measures the mass of a 1-carat diamond and finds it to be 200 milligrams, measured to the

vlada-n [284]

Answer:

$D=3.5\ gr/cm^3$

Step-by-step explanation:

Density

The density of an object of mass m and volume V is given by

$\displaystyle D=\frac{m}{V}$

It can be expressed in common units like $gr/lt, Kg/m^3, gr/cm^3$ or any other combination of proper mass [M] by volume [V] units.

The data provided in the question is

$m=200\ mg = 0.2\ gr$

$V=0.057\ cm^3$

Thus, the measured density is

$\displaystyle D=\frac{0.2\ gr}{0.057 \ cm^3}$

$\boxed{D=3.5\ gr/cm^3}$

We have expressed the result with 1 decimal place because the mass was measured to the nearest hundred milligrams (or one-tenth grams). Any further decimal is senseless because that precision comes from calculations, not from measurements.

6 0

3 years ago

Solve the inequality 3x - 5

SSSSS [86.1K]

The anwser would be -15
3 x (-5) = -15

4 0

4 years ago

The half-life of the isotope Osmium-183 is 12 hours. Choose the equation below that gives the remaining mass of Osmium-183 in gr

raketka [301]

The given equations are incomprehensible, I'm afraid...

You're given that osmium-183 has a half-life of 12 hours, so for some initial mass M₀, the mass after 12 hours is half that:

1/2 M₀ = M₀ exp(12k)

for some decay constant k. Solve for this k :

1/2 = exp(12k)

ln(1/2) = 12k

k = 1/12 ln(1/2) = - ln(2)/12

Now for some starting mass M₀, the mass M remaining after time t is given by

M = M₀ exp(kt )

So if M₀ = 590 g and t = 36 h, plugging these into the equation with the previously determined value of k gives

M = 590 exp(36k) = 73.75

so 73.75 ≈ 74 g of Os-183 are left.

Alternatively, notice that the given time period of 36 hours is simply 3 times the half-life of 12 hours, so 1/2³ = 1/8 of the starting amount of Os-183 is left:

590/8 = 73.75 ≈ 74

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3 years ago

A simulation of solar activity based on NASA studies shows that the sun experiences a major solar flare once every 12 years. Abo

Sunny_sXe [5.5K]

8 flares are most likely to occur in one century. However, the answer is 8.33, so take that into account.

7 0

3 years ago