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3 years ago

1/6-1/8 subtract and simplify

Mathematics

1 answer:

Alex Ar [27]3 years ago

4 0

find common denominator

8/48-6/48=

2/48=

1/24 is your answer.

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Find the prime factorization of 50 using exponents

Elden [556K]

Five squared with the little two and 2 because 5 Times 10 is 50 and five is prime so then from 10 and you get five and two and both five and two are prime

5 0

3 years ago

30 POINTS AVAILABLE

kherson [118]

Answer:

$\large\boxed{(x-2)^2+(y-1)^2=34}$

Step-by-step explanation:

The equation of a circle in standard form:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

We have the endpoints of the diameter: (-1, 6) and (5, -4).

Midpoint of diameter is a center of a circle.

The formula of a midpoint:

$\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)$

Substitute:

$h=\dfrac{-1+5}{2}=\dfrac{4}{2}=2\\\\k=\dfrac{6+(-4)}{2}=\dfrac{2}{2}=1$

The center is in (2, 1).

The radius length is equal to the distance between the center of the circle and the endpoint of the diameter.

The formula of a distance between two points:

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

Substitute the coordinates of the points (2, 1) and (5, -4):

$r=\sqrt{(5-2)^2+(-4-1)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}$

Finally we have:

$(x-2)^2+(y-1)^2=(\sqrt{34})^2$

3 0

3 years ago

Helppp im not sure about this one

PSYCHO15rus [73]

There’s no question here!

8 0

3 years ago

Consider the following ordered data. 6 9 9 10 11 11 12 13 14 (a) Find the low, Q1, median, Q3, and high. low Q1 median Q3 high (

IrinaVladis [17]

Answer:

Low Q1 Median Q3 High

6 9 11 12.5 14

The interquartile range = 3.5

Step-by-step explanation:

Given that:

Consider the following ordered data. 6 9 9 10 11 11 12 13 14

From the above dataset, the highest value = 14 and the lowest value = 6

The median is the middle number = 11

For Q1, i.e the median of the lower half

we have the ordered data = 6, 9, 9, 10

here , we have to values as the middle number , n order to determine the median, the mean will be the mean average of the two middle numbers.

i.e

median = $\dfrac{9+9}{2}$

median = $\dfrac{18}{2}$

median = 9

Q3, i.e median of the upper half

we have the ordered data = 11 12 13 14

The same use case is applicable here.

Median = $\dfrac{12+13}{2}$

Median = $\dfrac{25}{2}$

Median = 12.5

Low Q1 Median Q3 High

6 9 11 12.5 14

The interquartile range = Q3 - Q1

The interquartile range = 12.5 - 9

The interquartile range = 3.5

7 0

3 years ago

Beginning inventory: $14,320 Merchandise Purchases: $7,880 Goods available for sale: (a) Ending inventory: $11,250 Cost of goods

VikaD [51]

Answer:

$22,200

$10,950

Step-by-step explanation:

Goods available for sale = beginning inventory + Merchandise Purchases

$14,320 + $7,880 = $22,200

Ending inventory = goods available for sale - ending inventory

$22,200 - $11,250 = $10,950

6 0

3 years ago