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irinina [24]
2 years ago
5

Solve the sequence:0,5,12,23,34____​

Mathematics
1 answer:
MrRissso [65]2 years ago
4 0

Step-by-step explanation:

the answer is 49

I may not be correct

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Sarah took the advertising department from her company on a round trip to meet with a potential client. Including Sarah a total
Fantom [35]

Answer:the number of coach tickets that she bought is 5

the number of first class tickets that she bought is 6

Step-by-step explanation:

Let x represent the number of coach tickets that she bought.

Let y represent the number of first class tickets that she bought.

The total number of people that went for the round trip is 11. This means that

x + y = 11

She was able to purchase coach tickets for ​$340 and first class tickets for ​$1190. She used her total budget for airfare for the​ trip, which was ​$8840. This means that

340x + 1190y = 8840 - - - - - - - 1

Substituting x = 11 - y into equation 1, it becomes

340(11-y) + 1190y = 8840

3740 - 340y + 1190y = 8840

- 340y + 1190y = 8840 - 3740

850y = 5100

y = 5100/850 = 6

x = 11 - y

x = 11 - 6 = 5

4 0
3 years ago
CHOOSE THE NUMBER STATEMENT BELOW THAT IS CORRECT
tiny-mole [99]

Answer:

what number statement

5 0
3 years ago
Find the measure of ∠. Show your work.
poizon [28]

Answer:

77

Step-by-step explanation:

If you have any questions about the way I solved,don't hesitate to ask

5 0
3 years ago
PLEASE HELP!!!!!!!!!!!!! DUE SOON
Sergio039 [100]

\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ h=-16t^2+\stackrel{\stackrel{v_o}{\downarrow }}{65}t


now, take a look at the picture below, so for 2) and 3) is the vertex of this quadratic equation, 2) is the y-coordinate and 3) the x-coordinate.


\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+65}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{65}{2(-16)}~~,~~0-\cfrac{65^2}{4(-16)} \right) \implies \left( \cfrac{65}{32}~,~0- \cfrac{4225}{-64}\right)


\bf \left( \cfrac{65}{32}~,~0+ \cfrac{4225}{64}\right)\implies \left( \stackrel{seconds}{2\frac{1}{32}}~~,~~ \stackrel{feet~hight}{66\frac{1}{64}}\right)

6 0
3 years ago
Thanks for the help ;)
Pani-rosa [81]

Answer:

A.

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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