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lara [203]
3 years ago
10

A gym has yoga classes each class has 14 students if there are c classes write an equation to represent the total number of stud

ents s taking yoga
Mathematics
1 answer:
Orlov [11]3 years ago
7 0

Answer:

14 times c = s

Step-by-step explanation:

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HELLLP ASAP
Kisachek [45]

Answer:

<u>1.</u> 20 boxes cost $36. The unit rate is 1:1.8 and when you multiply 1.8 by 5 you get 9. If you also multiply 1.8 by 20 you get 36.

<u>2.</u>He completes 12.5 pages in 200 minutes.

200 divided by 32

multiply that answer by 2

which is 12.5.

hope it helps!

3 0
3 years ago
Read 2 more answers
Nicola is multiplying 1/6 by 3/2<br> Use the drop-down menu to complete the statement
Over [174]
Greater than
1/6 x 3/2 = 1/4
1/4 > 1/6
5 0
3 years ago
What is four fifths plus eight ninths
Jobisdone [24]
4/5 + 8/9. First we have to find a common denominator for \frac{4}{5} and \frac{8}{9}. \\  \frac{4*9}{5*9}= \frac{36}{45} \ and \  \frac{8*5}{9*5}= \frac{40}{45} \\ Now we are going to add:\frac{36}{45}+ \frac{40}{45}= \frac{76}{45} Make \frac{76}{45} a mixed number:\frac{76}{45}=1 \frac{31}{45} So the answer is \boxed{1 \frac{31}{45}}
6 0
4 years ago
If Emily buys a bag of food and toys for her bunny and she buys o bag of food for 20 dollars and she buys 2 toys for 3 dollars e
sasho [114]

Answer:

23

Step-by-step explanation:


7 0
4 years ago
Suppose a punter kicks a football so that the upward component of its velocity is 80 feet per second. If the ball is 3 feet off
Alexxandr [17]
So hmm check the picture below

\bf \qquad \textit{initial velocity}\\\\ h = -16t^2+v_ot+h_o \qquad \text{in feet}\\&#10;\\&#10;v_o=\textit{initial velocity of the object}\\&#10;h_o=\textit{initial height of the object}\\&#10;h=\textit{height of the object at "t" seconds}\\\\&#10;-----------------------------\\\\&#10;h(t)=3+80t-16t^2\iff h(t)=-16t^2+80t+3

a)

well, clearly is 80 ft/s

b)

when t = 1? well 80(1)

c)

in the picture, x-axis is the time and y-axis is the height
so, it reaches its maximum at the vertex, after "x" seconds

\bf \begin{array}{lcccll}&#10;h(t)=&-16t^2&+80t&+3\\&#10;&\uparrow &\uparrow &\uparrow \\&#10;&a&b&c&#10;\end{array}\qquad &#10;\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad  {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)

so it reached the vertex after \bf -\cfrac{{{ b}}}{2{{ a}}}\quad seconds

d)

the maximum height of the ball is \bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\quad feet

8 0
3 years ago
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