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pogonyaev
4 years ago
7

Rain fell on four days in June. These amounts were recorded: 5 mm, 9 mm, 2 cm, and 2.6 cm. What was the total rainfall for June?

Mathematics
1 answer:
ivolga24 [154]4 years ago
8 0
Hello there

The correct answer in mm would be 60 mm

1 cm = 10 mm

So it would 5+9+20+26 = 60

In cm it would be 6 cm

10 mm = 1 cm

.5+.9+2+2.6 = 6

I hope this helped ^^
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Semmy [17]

Answer:

\boxed{x = 7}

Step-by-step explanation:

75 = 11x-2 <u><em>(Corresponding angles are equal)</em></u>

11x -2 = 75

Adding 2 to both sides

11x = 75+2

11x = 77

Dividing both sides by 11

x = 7

5 0
3 years ago
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Robert and Elaine ran around the high school track during gym class. robert ran 1/2 of the way around the track in 5/6 minute. L
jok3333 [9.3K]

<u>Answer</u>


Incorrect


<u>Explanation</u>

Unpack the problem:

Let the distance round the track to be X.

Speed is the ratio of distance to time.

Robert run a distance of (1/2)x

Elaine run a distance of (3/4)x


Make a plan:

Finding the speed of each.

Compare their speeds to determine who ran faster than who.


Solution:

Robert's speed =(1/2)x/(5/6)

=1/2×6/5x

= (3/5)x

= 0.6x

Elaine's speed = (3/4)x/(9/10)

= (3/4)×(10/9)x

= (5/6)x

= 0.83333x



<em>Elaine ran faster than Robert. </em>


<u>Look back and explain:</u>

0.83333x > 0.6x

Elaine's speed is higher than Robert's speed.

This shows that Elaine ran faster than Robert.

5 0
4 years ago
1
nadya68 [22]

Answer:

64

Step-by-step explanation:

2, 6, 28, 23,7.64.

because 64 is the furthest away from any other number

5 0
3 years ago
A cooler has a temperature of 32 degrees Fahrenheit. A bottled drink is placed in the cooler with an initial temperature of
FinnZ [79.3K]

Answer:

k \approx 0.44

Step-by-step explanation:

Given function:

f(t) = (ce)^{-kt}+32

As per question statement:

Initial temperature of bottle is 70 ^\circ F.

i.e. when time = 0 minutes, f(t) = 70 ^\circ F

70 = ce^{-k\times 0}+32\\\Rightarrow 38 = ce^{0}\\\Rightarrow c = 38

After t = 3, f(t) = 42^\circ F

42 = 38 \times e^{-k\times 3}+32\\\Rightarrow 42-32 = 38 \times e^{-3k} \\\Rightarrow 10  = 38 \times e^{-3k} \\\Rightarrow e^{3k} = \dfrac{38}{10}\\\Rightarrow e^{3k} = 3.8\\\\\text{Taking } log_e \text{both the sides:}\\\\\Rightarrow log_e {e^3k} = log_e {3.8}\\\Rightarrow 3k \times log_ee=log_e {3.8}   (\because log_pq^r=r \times log_pq)\\\Rightarrow 3k \times 1=log_e {3.8}\\\Rightarrow 3k = 1.34\\\Rightarrow k = \dfrac{1.34}{3}\\\Rightarrow k \approx 0.44

Hence, the value is:

k \approx 0.44

8 0
4 years ago
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Sav [38]

Answer:

c.

Step-by-step explanation:

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3 years ago
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