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Umnica [9.8K]
3 years ago
15

Please I need help!!!!!!!!

Mathematics
2 answers:
seropon [69]3 years ago
7 0

Answer:

10 is the correct answer

nexus9112 [7]3 years ago
5 0

Answer:

Go with the third option 10!

i hope this helped!

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...help please I would appreciate it<br> No link please
suter [353]

Answer:

B

Step-by-step explanation:

7 0
3 years ago
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Sabra went for a long hike and burned 845 calories in 3 1/4 hours.Nelson decided to go for a bike ride. He burned 1,435 calories
sergey [27]

<u><em>Answer:</em></u>

Nelson burned the most calories per hour

<u><em>Explanation:</em></u>

To solve this question, we will get the amount calories burned by each in one hour and then compare the two values

To do this, we will divide the total amount of calories burned by the total time

<u>1- For Sabra:</u>

We are given that she burnt 845 calories in 3\frac{1}{25} (which is equivalent to 3.25) hours

<u>Therefore:</u>

Calories burnt in an hour = \frac{845}{3.25}=260 calories/hour

<u>2- For Nelson:</u>

We are given that he burnt 1435 calories in 4\frac{7}{8} (which is equivalent to 4.875) hours

<u>Therefore:</u>

Calories burnt in an hour = \frac{1435}{4.875}=294.36 calories/hour

<u>3- Comparing the two values:</u>

From the above calculations, <u>we can deduce that</u> Nelson burned the most calories per hour

Hope this helps :)

3 0
3 years ago
Carl drove from his house to work at an average speed of 35 miles per hour. The drive took him 25 minutes. If the
Yanka [14]
Carls speed going home is 1.2km an hour
8 0
3 years ago
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. Mr. Wayne has a 3-year contract for his cell phone service. He pays $124.65 each month to cover everyone in his family. How mu
telo118 [61]

Answer:

$4,523.40

Step-by-step explanation:

The cell phone cost is $124.65 per month. With 12 months in a year, we multiply $124.65 x 12 to get the answer $1,507.80. Since one year equals $1,507.80, we multiply this answer by 3 for 3 years to get $4,523.40. Thus three years of service with $124.65 a month would equal $4,523.40.

8 0
3 years ago
Find an equation of the line that passes through the points (-5, -3) and (3, 1)
Mkey [24]

\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{-3})\quad (\stackrel{x_2}{3}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{(-3)}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{(-5)}}}\implies \cfrac{1+3}{3+5}\implies \cfrac{4}{8}\implies \cfrac{1}{2}

\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{\cfrac{1}{2}}[x-\stackrel{x_1}{(-5)}]\implies y+3=\cfrac{1}{2}(x+5) \\\\\\ y+3=\cfrac{1}{2}x+\cfrac{5}{2}\implies y=\cfrac{1}{2}x+\cfrac{5}{2}-3\implies y = \cfrac{1}{2}x-\cfrac{1}{2}

6 0
3 years ago
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