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3 years ago

One of the factors of 1331x to the 3rd power-8y to the 3rd power.

Mathematics

1 answer:

artcher [175]3 years ago

6 0

Remember
a^3-b^3=(a-b)(a^2+ab+b^2)

(11x)³-(2y)³=(11x-2y)(121x²+22xy-4y²)

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likoan [24]

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$\frac{3}{8}$

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The graphs of the functions f(x)=|x-3| + 1 and G(x)=2x + 1 are drawn. which statement about these functions is true?

Roman55 [17]

The only true statement that compares the two functions is:

B: The solution to f(x)=g(x) is 1

<h3>Which statements are correct?</h3>

Here we have the functions:

f(x) = |x - 3| + 1

g(x) = 2x + 1

First, let's find the solution for:

f(x) = g(x)

|x - 3| + 1 = 2x + 1

|x - 3| = 2x

Notice that we have two options, x = 3:

|3 - 3| = 2*3

0 = 6 (x = 3 is not a solution)

And x = 1:

|1 - 3| = 2*1

2 = 2 (x = 1 is a solution).

Now to get the y-value where the graphs intersect, we just evaluate one of the functions in the solution we found above;

f(1) = |1 - 3| + 1 = 2 + 1 = 3

The graphs intersect when y = 3.

Then we conclude that the only true statement is:

B: The solution to f(x)=g(x) is 1

If you want to learn more about systems of equations, you can read:

brainly.com/question/13729904

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2 years ago

A box of 42 chocolates contains milk and dark chocolates in the ratio of 3:4. How many chocolates of each type there are?

Elza [17]

<h2>Answer:The number of milk chocolates are 18 and the number of dark chocolates are 24.</h2>

Step-by-step explanation:

Let the number of milk chocolates be $m$ .

Let the number of dark chocolates be $d$ .

Given that the box contains the milk and dark chocolates in the ratio $3:4$ .

So, $\frac{m}{d}=\frac{3}{4}$ ...(i)

Given that the total number of chocolates are $42$ .

So, $m+d=42$ . ...(ii)

Using (i) and (ii),

$m+\frac{4}{3}m=42$

$\frac{7}{3}m=42$

$m=18$

$d=\frac{4}{3}m=\frac{4}{3}\times 18=24$

So,the number of milk chocolates are $18$ and the number of dark chocolates are $24$ .

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ss7ja [257]

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