What is the answer for Expanding <br> 2t(6t-3)

8x^2y^3+6x^2y<br> Please factorize

stepladder [879]

Answer:

8x^2y^3+6x^2y

8x²y³+6x²y

2x²y(4y²+3)

5 0

3 years ago

An average person in the United States throws away 4.5 × 10² grams of trash per day. In 2013, there were about 3.2×106 people in

enyata [817]

Answer:

Since there are 4.5*10^2 grams of trash per day per person, we multiply that by 3.2*10^6 for everyone in the US and then multiply that by 365 to get how much in a year, which is 525600000000 or 5.256*10^11 (I think it's 11)

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Hannah has an offer from a credit carf issue foe 0% apr for the first 30 days and 12.22% apr afterwards, compounded daily. What

Alenkasestr [34]

DESIGNED TO INCREASE YOUR PROFIT MARGIN A highly specialized shovel machine developed for modern highspeed loading and product movement requirements. Mobility of big rubber wheels makes the Kimco Hough JH-30B Payloader an efficiency expert in agility. It reaches and goes around obstructions where track-type loaders cannot. Features a simple rugged boom with extra reach for maximum digging and breakout. Full reversing, "non-stop" power-shift transmission. Choice of gasoline or diesel power. There are two sizes in the Kimco Hough Payloader line - the JH-30B - capacity 3500 pounds and the JH6, capacity 5500 INTERNATIONAL pounds. There you go you can do it

6 0

3 years ago

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$