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nasty-shy [4]
3 years ago
6

during normal sleep, a bears heart beats about 50 times a minute. in its deepest state of hibernation a bears heart may beat 50

times in 6 minutes. during deep hibernation, how many times would the bears heart beat in 45 minutes?
Mathematics
2 answers:
Brilliant_brown [7]3 years ago
6 0

Answer:

375 times

Step-by-step explanation:

I used cross multiplication.

50 times=6 minutes

??? =45 minutes

50•45

_______=375

6

Stels [109]3 years ago
5 0

The bear beats its heart 50/6 times a minute during deep hibernation.

Thats 12 times a minute

45minutes x 12 times a minute

45x12 = 540

Answer: 540 times

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The cost of company 1: 10x+50
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3 years ago
In ΔKLM, the measure of ∠M=90°, the measure of ∠K=78°, and KL = 35 feet. Find the length of MK to the nearest tenth of a foot.
Margaret [11]

Answer:

7.3 feet

Step-by-step explanation:

osK=

hypotenuse

adjacent

​

=

35

x

​

\cos 78=\frac{x}{35}

cos78=

35

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​

35\cos 78=x

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Cross multiply.

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Type into calculator and round to the nearest tenth of a foot.

4 0
3 years ago
Read 2 more answers
ILL GIVE BRAINELST
Angelina_Jolie [31]

Answer: number 3 4 5

Step-by-step explanation:hope this helps!

5 0
3 years ago
A skier has decided that on each trip down a slope, she will do 3 more jumps than before. On her first trip she did 5 jumps. Der
taurus [48]
Since we are already given the amount of jumps from the first trial, and how much it should be increased by on each succeeding trial, we can already solve for the amount of jumps from the first through tenth trials. Starting from 5 and adding 3 each time, we get: 5 8 (11) 14 17 20 23 26 29 32, with 11 being the third trial.

Having been provided 2 different sigma notations, which I assume are choices to the question, we can substitute the initial value to see if it does match the result of the 3rd trial which we obtained by manual adding.

Let us try it below:

Sigma notation 1:

  10
<span>   Σ (2i + 3)
</span>i = 3

@ i = 3

2(3) + 3
12

The first sigma notation does not have the same result, so we move on to the next.

  10
<span>   Σ (3i + 2)
</span><span>i = 3
</span>
When i = 3; <span>3(3) + 2 = 11. (OK)
</span>
Since the 3rd trial is a match, we test it with the other values for the 4th through 10th trials.

When i = 4; <span>3(4) + 2 = 14. (OK)
</span>When i = 5; <span>3(5) + 2 = 17. (OK)
</span>When i = 6; <span>3(6) + 2 = 20. (OK)
</span>When i = 7; 3(7) + 2 = 23. (OK)
When i = 8; <span>3(8) + 2 = 26. (OK)
</span>When i = 9; <span>3(9) + 2 = 29. (OK)
</span>When i = 10; <span>3(10) + 2 = 32. (OK)

Adding the results from her 3rd through 10th trials: </span><span>11 + 14 + 17 + 20 + 23 + 26 + 29 + 32 = 172.
</span>
Therefore, the total jumps she had made from her third to tenth trips is 172.


3 0
3 years ago
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