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3 years ago

The point is on the x-axis and 12 units to the left of the y-axis.

Mathematics

1 answer:

tekilochka [14]3 years ago

7 0

Check the picture below.

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Find the quotient of 21x^2y^6+6xy^3-30xy/3xy.

kaheart [24]

$\bf ~~~~~~~~~~~~\textit{negative exponents}
\\\\
a^{-n} \implies \cfrac{1}{a^n}
\qquad \qquad
\cfrac{1}{a^n}\implies a^{-n}
\qquad \qquad 
a^n\implies \cfrac{1}{a^{-n}}
\\\\
-------------------------------$

$\bf \cfrac{21x^2y^6+6xy^3-30xy}{3xy}\implies \stackrel{\textit{distributing the denominator}}{\cfrac{21x^2y^6}{3xy}+\cfrac{6xy^3}{3xy}-\cfrac{30xy}{3xy}}
\\\\\\
\cfrac{21}{3}\cdot x^2x^{-1}y^6y^{-1}+\cfrac{6}{3}\cdot x^1x^{-1}y^3y^{-1}-\cfrac{30}{3}\cdot x^1x^{-1}y^1y^{-1}
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7x^{2-1}y^{6-1}+2x^{1-1}y^{3-1}-10x^{1-1}y^{1-1}\implies 7xy^5+2y^2-10$

3 0

3 years ago

Read 2 more answers

The cost of renting a pickup truck at Trucks Plus is $25 plus $15 cents per mile driven. The cost of renting a truck at Need-A-T

kodGreya [7K]

Okay so. Here's what it looks like : TRUCK PLUS: $25 is the fee upfront. so no matter what your GOING to pay $25 for using the company. Next 15 cents for every mile driven. So if the truck were to drive 3 miles it would be $.45 cents

NEED-A-TRUCK : Same thing. You have to pay $25 no matter what. But for this company you have to pay $.10 per mile. COST PER MILE means that basically the more miles you drive, the more cents you have to pay. Does that make sense? <span />

6 0

3 years ago

John made this model to show \frac{4}{7}\times\frac{13}{9} 7 4 × 9 13 Using John's model, what is \frac{4}{7}\times\frac{13}

harkovskaia [24]

Answer:

$\frac{4}{7}\times\frac{13}{9} = \frac{52}{63}$

Step-by-step explanation:

Given

See attachment for model

Required

Determine $\frac{4}{7}\times\frac{13}{9}$ from the model

The model is represented by:

$\frac{4}{7}\times\frac{13}{9} = \frac{4}{7}\times\frac{9}{9} + \frac{4}{7}\times\frac{4}{9}$

To get: $\frac{4}{7}\times\frac{9}{9}$ , we consider the first partition

The number of shaded box is 63 ---- this represents the denominator

The total boxes shaded at the bottom is 36 ---- this represents the numerator

So, we have:

$\frac{4}{7}\times\frac{9}{9} = \frac{36}{63}$

To get: $\frac{4}{7}\times\frac{9}{9}$ , we consider the first partition

The number of shaded box is 63 ---- this represents the denominator

The total boxes shaded at the bottom is 16 (do not count the gray boxes) ---- this represents the numerator

So, we have:

$\frac{4}{7}\times\frac{4}{9} =\frac{16}{63}$

The equation becomes:

$\frac{4}{7}\times\frac{13}{9} = \frac{4}{7}\times\frac{9}{9} + \frac{4}{7}\times\frac{4}{9}$

$\frac{4}{7}\times\frac{13}{9} = \frac{36}{63} + \frac{16}{63}$

$\frac{4}{7}\times\frac{13}{9} = \frac{36+16}{63}$

$\frac{4}{7}\times\frac{13}{9} = \frac{52}{63}$

5 0

3 years ago

Expand ( x - 1/x^2)^4

emmainna [20.7K]

Answer:

We want to expand the expression:

$(x - \frac{1}{x^2} )^4$

We can just do it by brute force, this is:

First, rewrite our expression as the product of two square factors:

$(x - \frac{1}{x^2} )^4 = (x - \frac{1}{x^2} )^2*(x - \frac{1}{x^2} )^2$

Now we can expand each one these two factors:

$(x - \frac{1}{x^2} )^2 = (x - \frac{1}{x^2} )*(x - \frac{1}{x^2} ) = x^2 + \frac{1}{x^4} -2*x*\frac{1}{x^2}$

That can be simplified to

$x^2 - \frac{2}{x} + \frac{1}{x^4}$

Now we can replace that in our original expression to get:

$(x^2 - \frac{2}{x} + \frac{1}{x^4})*(x^2 - \frac{2}{x} + \frac{1}{x^4})$

Now we can expand that last product, to get:

$(x^2)^2 + 2*(x^2)*(-\frac{2}{x} ) + 2*(x^2)*(\frac{1}{x^4}) + 2*(\frac{-2}{x})*(\frac{1}{x^4}) + (\frac{-2}{x} )^2 + (\frac{1}{x^4})^2$

We can simplify that to:

$x^4 - 4x + 2x^2 - \frac{4}{x^5} + \frac{4}{x^2} + \frac{1}{x^8}$

That is the expanded expression.

6 0

3 years ago

Please tell me the answer! I'm crying in confusion! :(

Oksi-84 [34.3K]

7RP: 1.3 sugar for one ounce. I came up with this by dividing 14 into 10 1/2

7NS: I'm just going to say -1/8. but I don't know for absolute sure, sorry

7EE: $65+$.80x=$125

7G: m<Q=150

7SP: There's a 50% chance that they will all land on heads.

7 0

3 years ago