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icang [17]
3 years ago
7

Identify the properties of the given quadratic y = −3x2 + 6x + 17

Mathematics
2 answers:
CaHeK987 [17]3 years ago
8 0

Answer:

a= -3 b=6 c=17 is the correct answer

Step-by-step explanation:

emmasim [6.3K]3 years ago
6 0

Answer:

after that its Axis of symmetry is 1,vertex is 1,20 and the vertex is at a maximum

Step-by-step explanation:

good luck with school

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Answer:

1.6 cups of sugar

Step-by-step explanation:

0.4 cups per batch.

4 batches.

Total sugar required = 0.4 x 4

= 1.6 cups of sugar

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A dog kennel charges $40 to board a dog for one night and $35 per night for each night after that. Henry paid a total of $215 fo
sveticcg [70]

Answer:

5 nights.

Step-by-step explanation:

215=35x+40

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2 years ago
Select all equations for which -3 is a solution.
ad-work [718]

Answer:

look it up

Step-by-step explanation:

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3 years ago
Kyle is saving his money to buy a computer for $300. He has already saved $75 and plans to save an additional $22.50 each day. W
vladimir1956 [14]

Answer:

  75 + 22.50d = 300

Step-by-step explanation:

The equation sums the amount Kyle currently has (75) with the amount he saves in d days (22.50d). It sets that total to 300, since Kyle hopes to save a total of $300 in d days.

7 0
3 years ago
The half-life of carbon 14 is years. How much would be left of an original -gram sample after 2,292 years? (To the nearest whole
lozanna [386]

Answer:

A = 0.75 gram or 1 gram

Step-by-step explanation:

The half-life of carbon 14 is years. How much would be left of an original -gram sample after 2,292 years? (To the nearest whole number).

We can use the following formula for half-life of ^{14}C to find out how much is left from the original sample after 2,292 years:  

A = A_{0}e^{-0.000124t}

where:

<em>A</em> is the amount left of an original gram sample after <em>t</em> years, and  

A_{0} is the amount present at time <em>t</em> = 0.

The half-life of  ^{14}C  is the time <em>t</em> at which the amount present is one-half the amount at time <em>t </em>= 0.

If 1 gram of  ^{14}C is present in a sample,  

 Solve for A when t = 2,292:  

Substituting A_{0}  = 1 gram into the decay equation, and we have:  

A = A_{0}e^{-0.000124t}

A = A_{0}e^{-0.000124(2,292)}

A = 0.75 g or 1 g  

6 0
2 years ago
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