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irga5000 [103]
3 years ago
10

Solve the following eqution, (4p+q)3

Mathematics
1 answer:
nirvana33 [79]3 years ago
8 0
(4p+q)3 is not an equation.  It doesn't ask any question,
and doesn't seek an answer.

It doesn't ask for the values of 'p' or 'q', and it can't be evaluated
without knowing their values.

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[-3 1/4] greater or less than [-4]
Afina-wow [57]

Step-by-step explanation:

-3¼ is greater than -4

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Answer these 2 questions plz
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Solve |x + 7| &lt; 6<br> its for homework
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3 0
3 years ago
To support a local senior citizens center, a student club sent a flyer home to the n students in the school. The flyer said, "Pl
Licemer1 [7]

Answer:

the dollar amount the club would have if they reached half of their goal T + 50

the dollar amount the club would have if every student at the school donated 50 cents to the cause 0,5T

the dollar amount the club could donate if they made $50 more than their goal 0.25n

the dollar amount the club would still need to raise to reach its goal after every student at the school donated 50 cents 0.5n

the dollar amount the club would have if half of the students at the school each gave 50 cents T - 0.5n

Step-by-step explanation:

the dollar amount the club would have if they reached half of their goal T + 50

the dollar amount the club would have if every student at the school donated 50 cents to the cause 0,5T

the dollar amount the club could donate if they made $50 more than their goal 0.25n

the dollar amount the club would still need to raise to reach its goal after every student at the school donated 50 cents 0.5n

the dollar amount the club would have if half of the students at the school each gave 50 cents T - 0.5n

3 0
3 years ago
A dose of a new pharmaceutical drug is decaying according to the half-life model f(t) = Ne-0.2174, where N is the dose
Rudiy27

Answer:

  14.7

Step-by-step explanation:

Perhaps your equation is ...

  f(t) = N·e^(-0.2174t)

We want f(6) = 4, so ...

  4 = N·e^(-0.2174·6) = N·e^-1.3044 ≈ 0.2713N

  N = 4/0.2713 = 14.7 . . . . milligrams

The dose amount was about 14.7 mg.

6 0
4 years ago
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