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2 years ago

Help! Stuck on this one!

Mathematics

1 answer:

gtnhenbr [62]2 years ago

4 0

so we know that each m&m weighs about 2g.

the next line says to write a direct variation equation.

(definition: mathematical relationship between two variables that can be expressed by an equation in which one variable is equal to a constant times the other)

we then make the equation ( w=xm), then we use that equation to figure out the next line, which is to find the amount of m&ms that would fit in a bag of 1000.

(my head cant really process part a cause of a recent headache, so sorry if its wrong)

the answer to part b would be 500, which you can provide evidence based off of your part a equation.

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A salesperson's commission rate is 5 %. What is the commission from the sale of 32,000 worth of furnaces? Use pencil and pape

Len [333]

Answer: Answer:

Its sad that you have to uses this app to get the right answers.

Have you ever stopped to ponder if someone is trying to scam/Catfish you?

Also I bet that you haven't even tried reading this question.

I know that you can do it yourself!

If not then, Sorry to hear.

Also

The answer is:

Do it yourself, Because if you try..You'll have a better answer then I ever could. Or anyone TBH

6 0

2 years ago

What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

$(2 x+y)^{5}$

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}$

$$+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}$

Let us solve the term one by one.

$$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}$

$$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y$

$$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}$

$$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}$

$$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}$

$$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}$

Substitute these into the above expansion.

$(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}$

The coefficient of $x^{2} y^{3}$ is 40.

Option C is the correct answer.

5 0

2 years ago

HELP ME PLEASE!!!

ser-zykov [4K]

parallel lines, have the same exact slope, hmmm what is the slope of y = -2/3 x + 1/3 anyway? well, low and behold, the equation is already in slope-intercept form, therefore

$\bf ~~y=\stackrel{slope}{-\cfrac{2}{3}}x+\cfrac{1}{3}~~$ has a slope of -2/3.

so we're really looking for a line whose slope is -2/3 and runs through 9,4.

$\bf (\stackrel{x_1}{9}~,~\stackrel{y_1}{4})~\hspace{10em}slope = m\implies -\cfrac{2}{3}\\\\\\\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-4=-\cfrac{2}{3}(x-9)\\\\\\y-4=-\cfrac{2}{3}x+6\implies y=-\cfrac{2}{3}x+10$