Apply the distributive to a factor out the greatest common factor 64 + 40

HELP ASAP. The two lines graphed on the grid each represent an equation.

sammy [17]

Answer:

B

Step-by-step explanation:

the lines cross at (-2,1)

3 0

3 years ago

What is the following sum?<br>(please show how you worked it out)

AleksAgata [21]

Answer:

$4\sqrt[3]{2}x(\sqrt[3]{y}+3xy\sqrt[3]{y} )$

Step-by-step explanation:

Let's start by breaking down each of the radicals:

$\sqrt[3]{16x^3y}$

Since we're dealing with a cube root, we'd like to pull as many perfect cubes out of the terms inside the radical as we can. We already have one obvious cube in the form of $x^3$ , and we can break 16 into the product 8 · 2. Since 8 is a cube root -- 2³, to be specific, we can reduce it down as we simplify the expression. Here our our steps then:

$\sqrt[3]{16x^3y}\\=\sqrt[3]{2\cdot8\cdot x^3\cdot y}\\=\sqrt[3]{2} \sqrt[3]{8} \sqrt[3]{x^3} \sqrt[3]{y} \\=\sqrt[3]{2} \cdot2x\cdot \sqrt[3]{y} \\=2x\sqrt[3]{2}\sqrt[3]{y}$

We can apply this same technique of "extracting cubes" to the second term:

$\sqrt[3]{54x^6y^5} \\=\sqrt[3]{2\cdot27\cdot (x^2)^3\cdot y^3\cdot y^2} \\=\sqrt[3]{2}\sqrt[3]{27} \sqrt[3]{(x^2)^3} \sqrt[3]{y^3} \sqrt[3]{y^2}\\=\sqrt[3]{2}\cdot 3\cdot x^2\cdot y \cdot \sqrt[3]{y^2} \\=3x^2y\sqrt[3]{2} \sqrt[3]{y}$

Replacing those two expressions in the parentheses leaves us with this monster:

$2(2x\sqrt[3]{2}\sqrt[3]{y})+4(3x^2y\sqrt[3]{2} \sqrt[3]{y})$

What can we do with this? It seems the only sensible thing is to look for terms to factor out, so let's do that. Both terms have the following factors in common:

$4, \sqrt[3]{2} , x$

We can factor those out to give us a final, simplified expression:

$4\sqrt[3]{2}x(\sqrt[3]{y}+3xy\sqrt[3]{y} )$

Not that this is the same sum as we had at the beginning; we've just extracted all of the cube roots that we could in order to rewrite it in a slightly cleaner form.

6 0

3 years ago

Whats the Area? how do i find it? In these specific questions?

just olya [345]

The formula for finding the area of a square is:
${s}^{2}$

The formula for finding the area of a triangle is:
$\frac{1}{2} bh$

So for the square the area would be:
${2x}^{2} \times {2x}^{2} = {4x}^{2}$

The area for the triangle would be:

$\frac{1}{2} {2c}^{3} \times 4c = {4c}^{2}$

7 0

3 years ago

13. John and Jane ate lunch at a local restaurant and wanted to divide the bill evenly. The

postnew [5]

John will receive $9.80 back

6 0

3 years ago

According to a survey done at your school, about 42% of all the female students participate in 2 sport seasons. You randomly ask

IRISSAK [1]

Answer: 0.353

Step-by-step explanation:

We know that 42% of all the female students participate in 2 sports this season. (58% is the percent that did not)

This means that if we select a random student, there is a 0.42 probability that the student did participate in 2 sports this season.

Other thing that i will use is that, if we have a group of N objects, and we want to create a group of K objects out of the N, the number of combinations is:

$c = \frac{N!}{(N - K)!*K!}$

Now, out of 5, we want to find the proability that, at least 3 of them, did participate in at least 2 sports this season.

The probability that all 5 of them participated, is equal to the product of the individual probabilities. (so N = 5, and K = 5)

P = (0.42^5)*1 = 0.013

Now, if only 4 of them did participate (N = 5, k = 4), the probability is equal to:

P = (0.42^4)*(0.58)*5!/4! = (0.42^4)*(0.58)*5 = 0.09

The probability for only 3 is: (N = 5 and K = 3)

P = (0.42^3)*(0.58^2)*5!/(3!*2!) = (0.42^3)*(0.58^2)*(5*4/2) = 0.25

Then the total probability is:

P = 0.013 + 0.09 + 0.25 = 0.353

7 0

3 years ago