What is the surface area of the cylinder?

Given the midpoint of (-10,7) and the endpoint of (5,-3), what is the other endpoint?

xxMikexx [17]

(-25,17)

Hope this helps.

4 0

3 years ago

Francisco and Ryan are stuck simplifying radical expressions. Francisco has simplified to the quantity of x to the one half powe

pogonyaev

Francisco <span>has simplified to the quantity of x to the one half power, over x to the three eighteenth power.
(x^1/2) / (x^3/18)
x^(1/2-3/18)
x^1/3

Ryan has simplified to the twenty seventh root of the quantity of x to the second times x to the third times x to the fourth.
</span>²⁷√x²x³x⁴
²⁷√x⁹
(x)^9/27
x^1/3

8 0

3 years ago

Read 2 more answers

Compare the sum of the mixed numbers to the sum of the decimals 5.815+6.021=11.836

babunello [35]

The decimals you have.

The mixed numbers would be 5 815/1000 + 6 21/1000 = 11 836/1000

4 0

3 years ago

4 times the quantity of a number decreased by one is the same as two more than the same number. Find the number

Yanka [14]

The hard part is just making sense of what they are asking for.

4 times the quantity of a number decreased by one is just the expression:

4 * (x-1)

Where "x" is our mystery number.

Two more than the same number is just:

x + 2

Where "x" is our same mystery number.

They say that these two expressions are the same, so just set them equal to each other.

4*(x-1) = x+2

4x - 4 = x +2

3x -4 = 2

3x = 6

x = 2

The mystery number is 2. Plug in 2 and you will see that it works!

5 0

3 years 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