The distance around a circular region is known as its_______

dangina [55]

you multiply the commission by her sales and get 25.20$ then add that to her bae pay of 280 a week and get 305.20$ as your answer.

4 0

2 years ago

Shaggy earned

Luba_88 [7]

Answer

x>21.2

Step-by-step explanation:

7.55x+100>160

-100 on both sides

7.55x>60

/ 7.55 on both sides

x>21.2

That is a bit of a high number, but im pretty sure that I did the math right

> is less than or equal to in this equation bc idk how to make the real sign

8 0

3 years ago

HELP! Evaluate.<br><br> 45 + 22 ÷ 11 • {[(60 – 15) • 60] + 25}

siniylev [52]

Step-by-step explanation: [(60 – 15) • 60] + 25} = 45.60 +25

=2725

45 + 22 ÷ 11. 2725

45 + 2 . 2725

45+ 5450

=5495

4 0

2 years ago

A salesperson's commission rate is 4%. What is the commission from the sale of $32,000 worth of furnaces? Use pencil and paper.

Delvig [45]

Answer:

Step-by-step explanation:

commission from sale of $32,000 = $32,000 x 4% = $1,280

Suppose sales would double, comission would be doubled too.

Assume sales be S and commission be C

C = S * 4% (they have a linear positive relation)

if S double to 2S, new commission = (2S)*4% = 2*(S*4%) = 2 (old commission)

8 0

1 year ago

have two one-quart jars; the first is filled with water, and the second is empty. I pour half of the water in the first jar into

vfiekz [6]

Answer:

water in quarts is in the first jar after 10th pour = 12/11

Step-by-step explanation:

Let X represent first jar and Y represents second jar.

have two one-quart jars; the first is filled with water, and the second is empty

Lets give the initial value of 2 to the first jar which is filled with water. Lets say there are two liters of water in first jar.

Lets give the initial value of 0 to the second as it is empty.

So before any pour, the values are:

X: 2

Y: 0

pour half of the water in the first jar into the second

After first pour the value of jar X becomes:

Previously it was 2.

Now half of water is taken i.e. half of 2

2 - 1 = 1

So X = 1

The value of jar Y becomes:

The half from jar X is added to second jar Y which was 0:

After first pour the value of jar Y becomes:

0 + 1 = 1

Y = 1

a third of the water in the second jar into the first

After second pour the value of jar X becomes:

Previously it was 1.

Now third of the water in second jar Y is added to jar X

1 + 1/3

= (3 + 1)/3

= 4/3

X = 4/3

After second pour the value of jar Y becomes:

Previously it was 1.

Now third of the water in Y jar is taken and added to jar X so,

1 - 1/3

= (3 - 1)/3

= 2/3

Y = 2/3

a fourth of the water in the first jar into the second

After third pour the value of jar X becomes:

Previously it was 4/3.

Now fourth of the water in the first jar X is taken and is added to jar Y

= 3/4 * (4/3)

= 1

X = 1

After third pour the value of jar Y becomes:

Previously it was 2/3

Now fourth of the water in the second jar X is added to jar Y

= 2/3 + 1/4*(4/3)

= 2/3 + 4/12

= 1

Y = 1

a fifth of the water in the second jar into the first

After fourth pour the value of jar X becomes:

Previously it was 1

Now fifth of the water in second jar Y is added to jar X

= 1 + 1/5*(1)

= 1 + 1/5

= (5+1) / 5

= 6/5

X = 6/5

After fourth pour the value of jar Y becomes:

Previously it was 1.

Now fifth of the water in Y jar is taken and added to jar X so,

= 1 - 1/5

= (5 - 1) / 5

= 4/5

Y = 4/5

a sixth of the water in the first jar into the second

After fifth pour the value of jar X becomes:

Previously it was 6/5

Now sixth of the water in the first jar X is taken and is added to jar Y

5/6 * (6/5)

= 1

X = 1

After fifth pour the value of jar Y becomes:

Previously it was 4/5

Now sixth of the water in the first jar X is taken and is added to jar Y

= 4/5 + 1/6 (6/5)

= 4/5 + 1/5

= (4+1) /5

= 5/5

= 1

Y = 1

a seventh of the water in the second jar into the first

After sixth pour the value of jar X becomes:

Previously it was 1

Now seventh of the water in second jar Y is added to jar X

= 1 + 1/7*(1)

= 1 + 1/7

= (7+1) / 7

= 8/7

X = 8/7

After sixth pour the value of jar Y becomes:

Previously it was 1.

Now seventh of the water in Y jar is taken and added to jar X so,

= 1 - 1/7

= (7-1) / 7

= 6/7

Y = 6/7

a eighth of the water in the first jar into the second

After seventh pour the value of jar X becomes:

Previously it was 8/7

Now eighth of the water in the first jar X is taken and is added to jar Y

7/8* (8/7)

= 1

X = 1

After seventh pour the value of jar Y becomes:

Previously it was 6/7

Now eighth of the water in the first jar X is taken and is added to jar Y

= 6/7 + 1/8 (8/7)

= 6/7 + 1/7

= 7/7

= 1

Y = 1

a ninth of the water in the second jar into the first

After eighth pour the value of jar X becomes:

Previously it was 1

Now ninth of the water in second jar Y is added to jar X

= 1 + 1/9*(1)

= 1 + 1/9

= (9+1) / 9

= 10/9

X = 10/9

After eighth pour the value of jar Y becomes:

Previously it was 1.

Now ninth of the water in Y jar is taken and added to jar X so,

= 1 - 1/9

= (9-1) / 9

= 8/9

Y = 8/9

a tenth of the water in the first jar into the second

After ninth pour the value of jar X becomes:

Previously it was 10/9

Now tenth of the water in the first jar X is taken and is added to jar Y

9/10* (10/9)

= 1

X = 1

After ninth pour the value of jar Y becomes:

Previously it was 8/9

Now tenth of the water in the first jar X is taken and is added to jar Y

= 8/9 + 1/10 (10/9)

= 8/9 + 1/9

= 9/9

= 1

Y = 1

a eleventh of the water in the second jar into the first

After tenth pour the value of jar X becomes:

Previously it was 1

Now eleventh of the water in second jar Y is added to jar X

= 1 + 1/11*(1)

= 1 + 1/11

= (11 + 1) / 11

= 12/11

X = 12/11

After tenth pour the value of jar Y becomes:

Previously it was 1.

Now eleventh of the water in Y jar is taken and added to jar X so,

= 1 - 1/11

= (11-1) / 11

= 10/11

Y = 10/11

3 0

3 years ago

The distance around a circular region is known as its________​

The distance around a circular region is known as its________