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nexus9112 [7]
2 years ago
11

Translate to an inequality. Use the variable x for next year's salary.

Mathematics
1 answer:
enot [183]2 years ago
7 0

x has to be greater or equal to 16,000

x ≥ 16,000

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The track team is trying to reduce their time for a relay race. First they reduce their time by 2.1 minutes. Then they are able
konstantin123 [22]

Answer: 15.7 minutes

Step-by-step explanation:

Let x be the time in the beginning (in minutes).

Given: The track team is trying to reduce their time for a relay race.

First they reduce their time by t_1=2.1 minutes.

Then they are able to reduce that time by t_2=10

If their final time is 3.96 minutes, then

x-t_1-t_2=3.6\\\Rightarrow\ x=3.6+t_1+t_2\\\Rightarrow\ x=3.6+2.1+10\\\Rightarrow\ x= 15.7

Hence, their beginning time was 15.7 minutes.

7 0
2 years ago
Fill in missing information to make the equality true: <br> (... +2a)2 = … +12ab2+4a2.
BigorU [14]

Answer:

I suppose it should be

6b² and (6b²)²

or

6b² and 36b^{4}

Step-by-step explanation:

(a+b)² = a² + 2ab + b²

therefore

(6b²+2a)² = (6b²)² + 12ab² + 4a²

3 0
2 years ago
Read 2 more answers
What is the common ratio of the geometric sequence whose second and fourth terms are 6 and 54, respectively?
Gnom [1K]
The common ratio is 3.
0 0
2 years ago
Read 2 more answers
A group of randomly selected members of Mothers' Club were asked how many kids they have. The table below shows the results A gr
LenKa [72]

Answer:

54

Step-by-step explanation:

Estimate=Population size × Sample proportion

population size =  120

sample proportion = (9)/(9+6+5) = 9/20 (check on Cymath as a resource)

estimate = 120 x 9/20 = 54

​

5 0
2 years ago
Read 2 more answers
ZE is the angle bisector of measure YEX and the perpendicular bisector of GF, GX is the angle bisector of measure YGZ and the pe
den301095 [7]

Answer:

C

Step-by-step explanation:

The center of inscribed circle into triangle is point of intersection of all interior angles of triangle.

The center of circumscribed circle over triabgle is point of intersection of perpendicular bisectors to the sides.

Circumscribed circle always passes through the vertices of the triangle.

Inscribed circle is always tangent to the triangle's sides.

In your case angles' bisectors and perpendicular bisectors intesect at one point, so point A is the center of inscribed circle and the center of corcumsribed circle. Thus, these circles pass through the points X, Y, Z and G, E, F, respectively.

7 0
2 years ago
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