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MAVERICK [17]
3 years ago
6

The mass of a tiger at a zoo is 135 kilograms. Randy's cat has a mass of 5,000 grams. How many times greater is the mass of the

tiger than the mass of Randy's cat?
Mathematics
2 answers:
vitfil [10]3 years ago
8 0
Randy's cat if 5 kilograms, so therefore the tiger's mass is 27 times larger than the mass of Randy's cat.
scZoUnD [109]3 years ago
6 0
Well you have to convert randy's cat's weight from grams to kilograms. 1000 grams equals 1 kilogram so 5000 grams equals 5 kilograms. Randy's cat weighs 5 kilograms. So 135 - 5 = 130 kilograms. The tiger is 130 kilograms greater than Randy's cat
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SVETLANKA909090 [29]
This <em>expression</em>, is a d. binomial

First term: 12xyz
Second term: -45

hope this helps
5 0
2 years ago
Figure 1 and figure 2 are two congruent parallelograms drawn on a coordinate grid as shown below:
LenKa [72]

Answer:

  (a)  Reflection across the y-axis, followed by translation 10 units down

Step-by-step explanation:

Figure 2 is not a reflection across the origin of Figure 1, so neither of the double reflections will map one to the other.

Reflection across the y-axis will put the bottom point at (5, 3). The bottom point on Figure 2 is at (5, -7), so has been translated down by 3-(-7) = 10 units.

Figure 1 is mapped to Figure 2 by reflection over the y-axis and translation down 10 units.

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2 years ago
What is the slope of the line given by the equation y = 12x?
FrozenT [24]

Answer:

up 12 over 1

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Newton's law of cooling is:
Mnenie [13.5K]

Answer:

t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr

So it would takes approximately 6.9 hours to reach 32 F.

Step-by-step explanation:

For this case we have the following differential equationÑ

\frac{du}{dt}= -k (u-T)

We can reorder the expression like this:

\frac{du}{u-T} = -k dt

We can use the substitution w = u-T and dw =du so then we have:

\frac{dw}{w} =-k dt

IF we integrate both sides we got:

ln |w| = -kt +C

If we apply exponential in both sides we got:

w = e^{-kt} *e^c

And if we replace w = u-T we got:

u(t)= T + C_1 e^{-kt}

We can also express the solution in the following terms:

u(t) = (T_i -T_{amb}) e^{kt} +T_{amb}

For this case we know that k =-0.15 hr since w ehave a cooloing, T_{i}= 70 F, T_{amb}=11F, we have this model:

u(t) = (70-11) e^{-0.15t} +11

And if we want that the temperature would be 32F we can solve for t like this:

32 = 59 e^{-0.15 t} +11

21=59 e^{-0.15 t}

\frac{21}{59} = e^{-0.15 t}

If we apply natural logs on both sides we got:

ln (\frac{21}{59}) =-0.15 t

t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr

So it would takes approximately 6.9 hours to reach 32 F.

7 0
3 years ago
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VARVARA [1.3K]
I guess maybe 5/35 because there are only 5 chances that you could get one of the 5 in the 35 people
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3 years ago
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