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Studentka2010 [4]

3 years ago

10

a business employs 100 workers at the start of the year, but then some employees quit or are fired. After that the business has

42 employees

2 answers:

Dennis_Churaev [7]3 years ago

8 0

100 - 42
= 58
I the question is how many got fired

BartSMP [9]3 years ago

5 0

52 workers or employees?

Your question is?

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What is the halfway between 5.3 and 5.4

murzikaleks [220]

Answer:

5.35

Step-by-step explanation:

5.3+5.4/2

10.7/2

5.35

5 0

3 years ago

2) Line segment MK has endpoints at (2, 3) and (5, ?4). Segment M'K' is the reflection of MK over the y-axis. Which statement de

Marianna [84]

Answer:

Option C -M'K' is the same length as MK

Step-by-step explanation:

Given : Line segment MK has endpoints at (2, 3) and (5,4)

M'K' is the reflection of MK over the y-axis

By definition of reflection: reflection of point (x,y) across the the y-axis is the point (-x,y)

which implies M'K' has end points (-2,3) and (-5,4)

Now, we find the length of MK

let $(x_1,y_1)=(2,3)\\\\(x_2,y_2)=(5,4)$

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

⇒ $d=\sqrt{(2-5)^2+(4-3)^2}$

⇒ $d=\sqrt{9+1}$

⇒ $d=\sqrt{10}$ ....(1)

Now, we find the length of M'K'

let $(x_1^{'},y_1^{'})=(-2,3)\\\\(x_2^{'},y_2^{'})=(-5,4)$

$d^{'}=\sqrt{(x_2^{'}-x_1^{'})^2+(y_2^{'}-y_1^{'})^2}$

⇒ $d^{'}=\sqrt{(-2+5)^2+(3-4)^2}$

⇒ $d^{'}=\sqrt{9+1}$

⇒ $d^{'}=\sqrt{10}$ .....(2)

from (1) and (2) we simply show that the length of MK and M'K' is equal

we can also refer the figure attached for reflection of MK and M'K'

therefore, Option C is correct

3 0

4 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%28-3%2C3%29%20%28-2%2C4%29" id="TexFormula1" title="(-3,3) (-2,4)" alt="(-3,3) (-2,4)" align=

Natali5045456 [20]

Answer:

sry I can't understand how to do this question

3 0

3 years ago

Help this is due in 2 minutes

Alika [10]

Answer:

The percentage change each year is 2%

Step-by-step explanation:

Generally, we can have an exponential equation written as;

f(t) = I( 1 + r)^t

where f(t) is the value at a certain time

I is the initial value at t = 0

r is the growth rate

t is the time we are looking at

Thus, we have it that;

1.02 = 1 + r

r = 1.02 -1

r = 0.02

So, let us write 0.02 as a percentage

We have this as 2/100 which is the same as 2%

The percentage change each year is 2%

5 0

3 years ago

Please please please help me! The question is in the picture! Please

quester [9]

Answer:

option 3

Step-by-step explanation:

$=\frac{1}{2}(2x-5) +5\\=x-\frac{5}{2} +5\\\\$

3 0

3 years ago

Read 2 more answers