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3 years ago

Which statement is true about centimeters and kilometers?

Mathematics

1 answer:

olga2289 [7]3 years ago

6 0

Answer:

Centimeters are smaller than kilometers

Step-by-step explanation:

Remember:

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A triangle has angles that measure 8xº,

romanna [79]

Answer:

8xº = 90º

6xº = 67.5º

2xº = 22.5º

x = 11.25°

Step-by-step explanation:

The sum of angles in a triangle is 180°

8xº + 6xº + 2xº = 180º

16xº = 180º

x = 11.25º

8xº = 11.25(8) = 90º

6xº = 67.5º

2xº = 22.5º

Hope this helps :)

3 0

3 years ago

19. Using clay, a student made a right circular cone of height 48cm and base

aev [14]

Answer:

fhgiignngmtotmtjfntktngigmtkgitmrjrnrkr28395027

Step-by-step explanation:

jejf ygm no fmfijfpnfimf

3 0

2 years ago

3( − 3) − 5 >− 3 − 6<br> (show your work)

egoroff_w [7]

Answer:

x > 3

Step-by-step explanation:

3x-9-5x > -3x - 6

3x-5x+3x > -6+9

x > 3

3 0

2 years ago

Please please please help

Elina [12.6K]

Median of (72 82 92 93 94 97 98 102)
Median: 93.5
MAD : 7.125

Median of ( 53 59 64 65 65 66 67 69)
Median : 65
MAD : 3.75

4 0

2 years ago

Which is the simplified form of the expression ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript nega

AlekseyPX

Answer:

The option "StartFraction 1 Over 3 Superscript 8" is correct

That is $\frac{1}{3^8}$ is correct answer

Therefore $[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}$

Step-by-step explanation:

Given expression is ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared

The given expression can be written as

$[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2$

To find the simplified form of the given expression :

$[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2$

$=(2^{-2})^{-3}(3^4)^{-3}\times (2^{-3})^2(3^2)^2$ ( using the property $(ab)^m=a^m.b^m$ )

$=(2^6)(3^{-12})\times (2^{-6})(3^4)$ ( using the property $(a^m)^n=a^{mn}$

$=(2^6)(2^{-6})(3^{-12})(3^4)$ ( combining the like powers )

$=2^{6-6}3^{-12+4}$ ( using the property $a^m.a^n=a^{m+n}$ )

$=2^03^{-8}$

$=\frac{1}{3^8}$ ( using the property $a^{-m}=\frac{1}{a^m}$ )

Therefore $[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}$

Therefore option "StartFraction 1 Over 3 Superscript 8" is correct

That is $\frac{1}{3^8}$ is correct answer

6 0

3 years ago

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