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2 years ago

The ratio of peter's height to albertina's height is 147cm what is peter's height

Mathematics

1 answer:

stellarik [79]2 years ago

8 0

Answer:

4'9.87

Step-by-step explanation:

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lesya692 [45]

Answer:

Option D. is the correct choice.

Step-by-step explanation:

Cylinder and a cone.

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3 years ago

△ABC is given with line m drawn through A parallel to BC¯¯¯¯¯¯¯¯. In the course of proving that the interior angle measures of △

aleksklad [387]

Answer:

I think a) and c is correct.

4 0

3 years ago

(GRAPH BELOW)

Rainbow [258]

Answer:

The tap drains approximately half the water from the tank in 15 minutes

The tank initially has 50 gallons of water

Step-by-step explanation:

The tap drains approximately half the water from the tank in 15 minutes. True.

This is true because see that the point (15,25) lies on the graph of the line.

Which means after 15 minutes the amount of water left in the tank is 25 gallons and 25 is half of 50.

The tap drains exactly one gallon from the tank every minute. False

This is false because the slope of the graph is $m=\frac{50-25}{0-15}=-\frac{5}{3}$

The tank initially has 30 gallons of water. False.

The y-intercept is the initial gallons of water which is 50.

The tank initially has 50 gallons of water. True.

The y-intercept is the initial gallons of water which is 50.

The tank takes 50 minutes to drain completely. False.

It is false because the x-intercept tells us the tank is drained completely and it is not 50 minutes but rather 30 minutes.

4 0

3 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

6 0

3 years ago

What is the answer to <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20" id="TexFormula1" title=" \frac{1}{2} " alt=

marta [7]

1. 5 -4x =6 +2x
+4x. +4x
5=6. +6x
-6 -6
-1 = 6x
--- ---
6. 6
-6=x
2. 9 - 2x = 7x
+2x. +2x
9 = 9x
--- ----
9. 9
1 = x

8 0

3 years ago

The ratio of peter's height to albertina's height is 147cm what is peter's height​

The ratio of peter's height to albertina's height is 147cm what is peter's height