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gizmo_the_mogwai [7]
4 years ago
10

Someone please help. Please explain how I should do it!

Mathematics
1 answer:
Ede4ka [16]4 years ago
7 0

Answer:

Step-by-step explanation:

if m and n are irrational number then the product of mn sometimes rational and sometime irrational

ex: √5 *√2=√10 irrational

ex: √8*√2=√16=4 rational

b-y=|x|+3

explain why |x|+3≥|x+3|

absolute value of any number is always positive

x           |x|+3          ≥            |x+3|

1              4              =               4             in this case equal

2             5                                5

-2             5             ≥                 1             in this case |x|+3≥|x+3|

in case of negative value of x then |x|+3>|x+3|

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Which statement is true about the length of a square with an area of 200 square inches?
erica [24]

Answer:

D

Step-by-step explanation:

the square root is between 14.1 and 14.2.

4 0
3 years ago
Consider the shape below. (1) determine the value of x so that the volume of the shape is 165cm cube, and (2) once you have foun
Reika [66]

Answer:

  • x = 5 cm
  • 248 cm²

Step-by-step explanation:

The volume of the solid with x=0 is the product of its length, width, and height:

  V0 = (10 cm)(3 cm)(8 cm) = 240 cm³

The volume of the cutout is the same:

  Vc = (x cm)(3 cm)(x cm) = 3x² cm³

Then the volume of the solid with the cutout is ...

  V1 = V0 -Vc = 240 cm³ - 3x² cm³

This is 165 cm³, so we have ...

  165 = 240 -3x²

  55 = 80 -x² . . . . . divide by 3

  x² = 80 -55 = 25 . . . . add x²-55

  x = √25 = 5 . . . . . take the square root

___

The surface area is the total of the front and back faces and the lateral area of all the faces between them. That lateral area is the product of the perimeter of the front (or back) face and the width (3 cm). The perimeter (p) is the total of all edge lengths around the front face:

  p = 10 + 8 + (10-x)/2 + x + x + x + (10-x)/2 + 8 = 36 +2x

For x=5, the perimeter is ...

  p = 36 +2·5 = 46

so the lateral area is ...

... la = (46 cm)(3 cm) = 138 cm².

The area of each of the front and back faces is the area of the overal rectangular shape (10 cm by 8 cm) less the area of the cutout (x² cm²). So, that face area is ...

  fa = 2×((10 cm)(8 cm) -(5 cm)²) = 2×(55 cm²) = 110 cm²

Then the total surface area is the sum of lateral area and face area:

  S = la + fa = 138 cm² + 110 cm² = 248 cm²

7 0
3 years ago
How to find it is related to the opposite angle in a cyclic quadrilateral is equal to 180 degeres
lorasvet [3.4K]

Answer: The measure of ∠BEF=110°.

Step-by-step explanation:

Given: In cyclic quadrilateral ABCD, ∠C = 110°

Then, ∠A+∠C=180°  [The sum of opposite angles in a cyclic quadrilateral is 180 degrees.]

⇒ ∠A+ 110°=180°

⇒ ∠A = 180°  - 110°

⇒ ∠A = 70°

In cyclic quadrilateral ABEF,

∠A+∠E=180°   [The sum of opposite angles in a cyclic quadrilateral is 180 degrees.]

⇒ 70° + ∠E=180°

⇒∠E=110°

Hence, the measure of ∠BEF=110°.

5 0
3 years ago
For each part, compare distributions (1) and (2) based on their medians and IQRs. You do not need to calculate these statistics;
umka2103 [35]

Answer:

a) The two distributions have the same median (6) but the 2nd distribution has a larger IQR (9.5 > 4).

b) The two distributions have a different median with distribution 2 having a bigger median (8 > 3), but they both have the same IQR (3).

c) The 2nd distribution has a slightly bigger median (7 > 6) and a slightly bigger IQR (4.5 > 4) than the 1st distribution.

d) The 2nd distribution consists of variables that are way bigger than those of the 1st distribution, hence, the median and IQR of the 2nd distribution are about ten folds of that of the 1st distribution.

Step-by-step explanation:

For any distribution,

The median is the variable at the middle when all the variables in the dsitribution are arranged in ascending or descending order.

The median is the (n+1)/2 th variabke.

where n = size of the distribution or number of variables. For these questions, the sample size is 5, hence, the median will always be the 3rd variable when the variables are arranged in ascending or descending order.

The IQR, known as the inter quartile range is a measure of dispersion for the distribution. Although, it isn't as effective as other measures of dispersion such as the standard deviation because the IQR unlike the the standard deviation isn't responsive to changes in the variables, especially ones at the end of the distribution.

The IQR is simply given mathematically as the third quartile minus the first quartile.

IQR = (Third quartile) - (First quartile)

Third quartile is the 3(n+1)/4 th variable. For a sample of n=5, the third quartile is the 4.5th variable, that is the average of the 4th and 5th variable.

First quartile is the (n+1)/4 th variable. For a sample of n=5, the first quartile is the 1.5th variable, that is the average of the 1st and 2nd variable.

Taking the questions one at a time

a) (1) 3,5,6,7,9

Median = 3rd variable = 6

Third quartile = (7+9)/2 = 8

First quartile = (3+5)/2 = 4

IQR = 8 - 4 = 4

(2) 3,5,6,7,20

Median = 3rd variable = 6

Third quartile = (7+20)/2 = 13.5

First quartile = (3+5)/2 = 4

IQR = 13.5 - 4 = 9.5

The two distributions have the same median (6) but the 2nd distribution has a larger IQR (9.5 > 4).

b) (1) 1,2,3,4,5

Median = 3rd variable = 3

Third quartile = (4+5)/2 = 4.5

First quartile = (1+2)/2 = 1.5

IQR = 4.5 - 1.5 = 3

(2) 6,7,8,9,10

Median = 3rd variable = 8

Third quartile = (9+10)/2 = 9.5

First quartile = (6+7)/2 = 6.5

IQR = 9.5 - 6.5 = 3

The two distributions have a different median with distribution 2 having a bigger median (8 > 3), but they both have the same IQR (3).

c) (1) 3,5,6,7,9

Median = 3rd variable = 6

Third quartile = (7+9)/2 = 8

First quartile = (3+5)/2 = 4

IQR = 8 - 4 = 4

(2) 3,5,7,8,9

Median = 3rd variable = 7

Third quartile = (8+9)/2 = 8.5

First quartile = (3+5)/2 = 4

IQR = 8.5 - 4 = 4.5

The 2nd distribution has a slightly bigger median (7 > 6) and a slightly bigger IQR (4.5 > 4) than the 1st distribution.

d) (1) 0,10,50,60,100

Median = 3rd variable = 50

Third quartile = (60+100)/2 = 80

First quartile = (0+10)/2 = 5

IQR = 80 - 5 = 75

(2) 0,100,500,600,1000

Median = 3rd variable = 500

Third quartile = (600+1000)/2 =800

First quartile = (0+100)/2 = 50

IQR = 800 - 50 = 750

The 2nd distribution consists of variables that are way bigger than those of the 1st distribution, hence, the median and IQR of the 2nd distribution are about ten folds of that of the 1st distribution.

Hope this Helps!!!

5 0
3 years ago
The table shows the cost of several items at a
Alex_Xolod [135]

Answer:

The answer is $10.50

Step-by-step explanation:

8 0
3 years ago
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