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hjlf
3 years ago
14

A school staff meeting had 72 teachers in attendance, 25% of whom were first-year teachers. How many first-year teachers were in

the meeting?
Mathematics
2 answers:
vaieri [72.5K]3 years ago
6 0

25% of whom were first year teachers out of 72 teachers in attendance in a school staff meeting equals 18 first-year teachers in the meeting. Therefore, there are 18 first-year teachers in the school staff meeting.

Norma-Jean [14]3 years ago
5 0

Answer:

18 first-year teachers were at the meeting

Step-by-step explanation:

.25 * 72 = 18

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Answer:

-85n - 7  = -92

Step-by-step explanation:

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The circle described by the equation <img src="https://tex.z-dn.net/?f=%28x-3%29%5E%7B2%7D%20%2B%28y%2B4%29%5E%7B2%7D%3D9" id="T
ElenaW [278]

Answer:

The point at which center of the image circle located is (-2, -3) so second option is correct.

Step-by-step explanation:

Given equation of circle is:

(x-3)^{2} + (y+4)^{2} = 9 (i)

The general equation of circle is mentioned below,

(x-h)^{2} + (y-k)^{2} = r^{2} (ii)

Here 'r' represents the radius of the circle and (h,k) shown the center of the circle.

By comparing equation (i) and equation (ii), we get

r^2 = 9

r = 3

(x-3)^{2} = (x-h)^{2}

(x-3) = (x-h)

h = 3

(y+4)^{2} = (y-k)^{2}

(y+4) = (y-k)

k = -4

So the center of given circle is (h,k) = (3,-4)

Also, the circle is translated 5 units left, that is towards the -x-axis. Therefore h = 3 - 5 = -2

Also, the circle is translated 1 unit up, that is towards the +y-axis. Therefore k = -4 + 1 = -3

Hence, the point at which center of the image circle located is (-2, -3) so second option is correct.

6 0
3 years ago
Read 2 more answers
Solve the compound inequality |3x-9|≤15 and |2x-3|≥5. Give answer in interval notation.
laila [671]

Answer:

The solution of |3x-9|≤15 is [-2;8] and the solution |2x-3|≥5 of is  (-∞,2] ∪ [8,∞)

Step-by-step explanation:

When solving absolute value inequalities, there are two cases to consider.

Case 1: The expression within the absolute value symbols is positive.

Case 2: The expression within the absolute value symbols is negative.

The solution is the intersection of the solutions of these two cases.

In other words, for any real numbers a and b,

  • if |a|> b then a>b or a<-b
  • if |a|< b then a<b or a>-b

So, being |3x-9|≤15

Solving: 3x-9 ≤ 15

3x ≤15 + 9

3x ≤24

x ≤24÷3

x≤8

or 3x-9 ≥ -15

3x ≥-15 +9

3x ≥-6

x ≥ (-6)÷3

x ≥ -2

The solution is made up of all the intervals that make the inequality true. Expressing the solution as an interval: [-2;8]

So, being |2x-3|≥5

Solving: 2x-3 ≥ 5

2x ≥ 5 + 3

2x ≥8

x ≥8÷2

x≥8

or 2x-3 ≤ -5

2x ≤-5 +3

2x ≤-2

x ≤ (-2)÷2

x ≤ -2

Expressing the solution as an interval: (-∞,2] ∪ [8,∞)

6 0
3 years ago
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