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2 years ago

Why is 1/6 greater than 1/8

2 answers:

7 0

1/6 is greater than 1/8, because when you change the fractions 1/6 will be 4/24 but 1/8 will be 3/24. To change the fractions you first have to find the lowest number that both denominators can go in, so it is 24. 6 can go into 24 four times, so you multiply the numerator by 4 making 1/6 4/24. 8 can go into 24 three times, so you multiply the numerator by 3 so 1/8 = 3/24.

expeople1 [14]2 years ago

4 0

Because 1/6 is closer to a whole and if you cut a pie into 8 slices and get 1 slice of the 8 it is smaller than 6 pieces cut and get 1 of the 6 and the pieces will be bigger

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Which inequality is false? A. 4 < 9 B. –6 < –3 C. –5 > –2 D. 12 > –7

CaHeK987 [17]

Simple...

which inequality is false?

A.) 4<9 (True)

B.) -6<-3 (True)

C.) -5>-2 (False)

D.) 12>-7 (True)

Thus, your answer.

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2 years ago

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Find the x-intercepts (if any) for the graph of the quadratic function.

OleMash [197]

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3 years ago

4. Express each percent as a decimal number and

il63 [147K]

A) 0.955 and 199/200
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1 year ago

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Photo above need help

Ira Lisetskai [31]

Answer:

m = 0

Step-by-step explanation:

<u>If two lines are parallel, their slopes are equal. </u>

The slope of x - axis is 0

(why?)

because, slope of a line is given by

$\boxed{\mathsf{slope\:of\:a\:line=tan \theta}}$

Θ is the angle made by the line with the x- axis.

Here, the line in question is the x- axis itself, and every line makes an angle of 0° with itself.

=> Θ = 0

=> tan Θ = 0

=> slope = 0

and the line in blue is parallel to the X- axis, therefore, it's slope is also 0

(also, a line makes an angle of 0° with the line it's parallel to)

$\mathfrak{\overline{There\:You\:Are!}}$

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2 years ago

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A rectangle has a height of 6k3 and a width of 2k2 + 4k +5.

Trava [24]

Answer: $12k^5+24k^4+30k^3$

Step-by-step explanation:

Given : The height of the rectangle = $6k^3$

The width of the rectangle = $2k^2+4k+5$

Formula : Area = height x width

Therefore , the area of triangle in terms of polynomial will be :

$6k^3\times( 2k^2+4k+5)\\\\= 6k^3(2k^2)+6k^3(4k)+6k^3(5)\ \ [\text{Using Distributive property}]\\\\=12k^{3+2}+24k^{3+1}+30k^3\ \ [\text{Using exponents rule}:\ a^n\times a^m=a^{n+m}]\\\\=12k^5+24k^4+30k^3$

Hence, the area of the entire rectangle = $12k^5+24k^4+30k^3$

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3 years ago