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3 years ago

Find the degree of the minomial. 6xy

Mathematics

2 answers:

kakasveta [241]3 years ago

5 0

The degree of the monomial 6xy is 2.

EastWind [94]3 years ago

4 0

Count up the exponents.

6xy

Both 'x' and 'y' have exponents of 1.

1 + 1 = 2

So this is a 2nd degree monomial.

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Rock musician Donny West is paid 15% on his CD sales and tour video sales. Last year, he sold one million CDs and 550,000 videos

olganol [36]

He had $5,000,000 in CD sales; $3,300,000 in video sales; and he earned $1,245,000 in royalties.

1,000,000 CDs * $5 each = $5,000,000
550,000 videos * $6 each = $3,300,000

Total sales = 5,000,000+3,300,000 = 8,300,000
15% royalties = 0.15(8,300,000) = 1,245,000

5 0

4 years ago

Which of the following are solutions to | x+3 = 4x - 7? Check all that apply.

just olya [345]

Answer:

x=5/4, 3/10

6 0

3 years ago

Use matrices to write the equation of the function in the form y = mx + b that contains the points

Fofino [41]

The line has a slope of -6 and a y intercept of 7.

The linear equation is given by:

y = mx + b;

where m is the slope of the line, b is the y intercept, y, x are variables.

The equation of the line passing through the points (6.-29) and (-4,31) is:

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\y-(-29)=\frac{31-(-29)}{-4-6}(x-6)\\\\y+29=-6(x-6)\\\\y=-6x+7$

Hence the line has a slope of -6 and a y intercept of 7.

Find out more at: brainly.com/question/13911928

3 0

2 years ago

A coin sold for $264 in 1976 and was sold again in 1989 for $ 459. Assume that the growth in the value V of the collector's it

Vaselesa [24]

K=195 is the answer

6 0

3 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