Calculate the equation of the line through the points (1, 1), (4,3), (10, 7) and (40, 27).

hram777 [196]

The question was incorrect. Please find the correct content below.

Compare 32/35 and 9/10.

Comparing 32/35 and 9/10 we have 32/35 is greater than 9/10, that is 32/35 > 9/10.

Fraction is the ratio of two numbers. The upper number is called Numerator and the Lower number is called the Denominator.

We know that if the denominators are the same for two fractions then which has the greatest numerator is a greater fraction than the other.

Given the fractions are 32/35, 9/10

To compare this two fractions we have to make denominators equal first.

LCM of 10,35 = 70

Calculating the fractions,

32/35 = (32*2)/(35*2) = 64/70

9/10 = (9*7)/(10*7) = 63/70

Since 64 > 63

So 64/70 > 63/70

Therefore, 32/35 > 9/10

Hence fraction 32/35 is greater than the other fraction 9/10.

Learn more about Fraction here -

brainly.com/question/78672

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6 0

1 year ago

You are considering a cylindrical container with a diameter of 14 inches and a height of 18 inches. Find the volume of this cont

Vsevolod [243]

Volume of a cylinder is (pi) times radius to the second power times height. 
Radius is half of the diameter, which would be 7.
Volume would be 2770.88

3 0

3 years ago

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$