All categories

3 years ago

What is the greatest common factor of 90 and 54?

Mathematics

2 answers:

Olegator [25]3 years ago

6 0

The greatest common factor of 54 and 90 is 18.

miskamm [114]3 years ago

6 0

To get the Greates Common Factor (GCF) of 90 and 54 we need to factor each value first and then we choose all the copies of factors and multiply them:

90: 2 3 3 5

54: 2 3 3 3

GCF: 2 3 3

The Greates Common Factor (GCF) is: 2 x 3 x 3 = 18

You might be interested in

HELP ASAP!! Sorry if it’s hard to see. The second answer options are, 4.15, 2.33, 17.2

nirvana33 [79]

Answer:

r = $\sqrt{\frac{A}{\pi } }$

r = 17.2 cm

Step-by-step explanation:

∵ The area of the circle A = πr²

∴ r² = A/π

∴ r = √(A/π)

∵ A = 54 cm²

∵ π = 22/7

∴ r = √(54/(22/7)) = √(54×7/22) = 17.2 cm

3 0

3 years ago

I donot understand what so ever but brainly has def been helping

faltersainse [42]

Answer:

what happened to you brooooooooo

4 0

2 years ago

Which polynomials are in standard form?

Kruka [31]

Answer:

Regular type is called the form in which polynomials are mostly used. The parameters are calibrated to the lowest degree. The polynomial x4 + 2x3+x+11 is standard, for instance, because four are most strong and three and one are the strongest.

Step-by-step explanation:

3 0

3 years ago

On a quiz there are four-mulptiple choice questions worth 3 points each and two true/false questions worth 1 point

Scilla [17]

Answer:

Step-by-step explanation:

7 0

2 years ago

Calculate the total area of the shaded region.

LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

$\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}$

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

$\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill$

$\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}$

7 0

2 years ago