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Studentka2010 [4]
3 years ago
14

Please help need an answer by today

Mathematics
1 answer:
love history [14]3 years ago
8 0
We don't know what the question is.

It could be "What is the selling price ?"
If that's the question, then . . .

Since the store marks up their cost by 60% before they sell
the thing, the price they sell it for is

                       (1.60 of their cost).

You know their cost.  The selling price is   (1.60  x  $80) .    
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A shirt is marked down 20% off the original price. If the original price was $24.00, what is the sale price of the shirt before
Marizza181 [45]

Answer:$19.20

Step-by-step explanation:

8 0
3 years ago
What is the factored form of x^12y^18+1
Black_prince [1.1K]

Answer:  The required factored form of the given expression is

(x^4y^6+1)(x^8y^{12}-x^4y^6+1).

Step-by-step explanation:  We are given to find the factored form of the following algebraic expression:

E=x^{12}y^{18}+1.

We will be using the following formula:

a^3+b^3=(a+b)(a^2+ab+b^2).

Now, we have

E\\\\=x^{12}y^{18}+1\\\\=(x^4y^6)^3+1^3\\\\=(x^4y^6+1)\{(x^4y^6)^2-x^4y^6\times1+1^2\}\\\\=(x^4y^6+1)(x^8y^{12}-x^4y^6+1).

Thus, the required factored form of the given expression is

(x^4y^6+1)(x^8y^{12}-x^4y^6+1).

4 0
3 years ago
Read 2 more answers
Billy wants to live in the area defined by y < 3x − 6. Explain how you can identify the houses in which Billy is interested i
Nadusha1986 [10]
Check the picture below.

to graph an inequality, you pretty much first off, need to graph the "equality" or equation, so for y < 3x − 6, first off, graph the line <span>y = 3x − 6.

and then you do a "true" or "false" check.

now, if you look at the line below, it splits the grid in two sections, so, let's check a point in either section, if one is false, the other is true and the other way around.

so, let's check the region on the left-hand-side, hmmmm say point 0,0, the origin.

y < 3x - 6

y = 0,  x = 0  thus

0 < 3(0) - 6

0 < -6   <--- now, is that really true? is 0 lesser than -6? not quite, is false.

recall that on the negative side, the closer to 0, the larger the value, so -1 is much larger than -1,000,000.

because the point 0,0 yielded a false inequality, that region is the "false region" and thus not shaded, therefore, the other side must be the "true region", and thus we shade that then.

we can run a quick check on that btw, say on point 4,4

y =4,   x = 4  thus

4 < 3(4) - 6

4 < 12 - 6

4 < 8    <---- yeap, is true, 4 is lesser/smaller than 8.

hmm ohh yes, the line needs to be dashed for a < or >, because it "has that boundary but it does not include it", unlike </span>⩽ and ⩾, which is a solid line, because it has that boundary and it includes it too.

3 0
3 years ago
the building on the poster is 36cm tall. is it possible to figure out how tall the building is on the photograph?if its possible
fiasKO [112]
No because the stupidity of you can not calculate
4 0
3 years ago
Plz help<br>urgent !!!!<br>will give the brainliest !!​
Mkey [24]

Answer:

X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}

Step-by-step explanation:

The question we have at hand is, in other words,

\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix} - where we have to solve for the value of X

If we have to isolate X here, then we would have to take the inverse of the following matrix ...

\begin{bmatrix}4&2\\ \:1&1\end{bmatrix} ... so that it should be as follows ... \begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}

Therefore, we can conclude that the equation as to solve for " X " will be the following,

X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix} - First find the 2 x 2 matrix inverse of the first portion,

\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1} = \frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}= \frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix} = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix} ,

X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix} - Simplifying this, we should get ...

\begin{bmatrix}1&3\\ 2&4\end{bmatrix} ... which is our solution.

7 0
3 years ago
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