Pls with step. plsss. is anyone solving this?pls help

Mathematics

1 answer:

PtichkaEL [24]3 years ago

8 0

Answer:

Step-by-step explanation:

12ab - 10b² - 18a² + 9ab + 12b² + 14a²

= 12ab + 9ab - 10b² + 12b² -18a² + 14a² {Combine like terms}

= (12 + 9)ab + (-10 + 12)b² + (-18 + 14)a²

= 21ab + 2b² - 4a²

ab + 2b² + 3b² - a² = ab + (2+3)b² - a²

= ab + 5b² - a²

Now find the difference

ab + 5b² - a² - [ 21ab + 2b² - 4a²]

= ab + 5b² -a² - 21ab - 2b² + 4a²

= ab - 21ab + 5b² - 2b² - a² + 4a²

= -20ab + 3b² + 3a²

2) 2x³ – 3x² y +2xy²+3y³ - unknown polynomial = x³ -2x² y+3xy²+4y³

2x³ – 3x² y +2xy²+3y³ = x³ -2x² y+3xy²+4y³ + unknown polynomial

2x³ – 3x² y +2xy²+3y³ - [ x³ -2x² y+3xy²+4y³] = unknown polynomial

2x³ – 3x² y +2xy²+3y³ - x³ + 2x² y- 3xy²- 4y³ =

2x³ - x³ -3x²y +2x²y + 2xy² - 3xy² + 3y³ - 4y³=

x³ - x²y - xy² - y³

Answer: x³ - x²y - xy² - y³

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gamit kalang po ng photo math para masolve na problem mo

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Find the three angles of the triangle formed using the position vectors 2i hat − j + 5k and i hat + 2j + 4k and the line segment

victus00 [196]

Given

$\vec{a}=2\vec{i}-\vec{j}+5\vec{k},$

$\vec{b}=\vec{i}+2\vec{j}+4\vec{k},$

you can find

$\vec{a}-\vec{b}=2\vec{i}-\vec{j}+5\vec{k}-(\vec{i}+2\vec{j}+4\vec{k})=\vec{i}-3\vec{j}+\vec{k}.$

Three vectors $\vec{a}, \vec{b}, \vec{a}-\vec{b}$ form a triangle.

$\cos\angle 1=\dfrac{\vec{a}\cdot \vec{b}}{|\vec{a}|\cdot |\vec{b}|}=\dfrac{2\cdot 1+(-1)\cdot 2+5\cdot 4}{\sqrt{2^2+(-1)^2+5^2}\cdot \sqrt{1^2+2^2+4^2} }=\dfrac{20}{\sqrt{30} \cdot \sqrt{21} }.$

$\cos\angle 2=\dfrac{\vec{a}\cdot (\vec{a}-\vec{b})}{|\vec{a}|\cdot |\vec{a}-\vec{b}|}=\dfrac{2\cdot 1+(-1)\cdot (-3)+5\cdot 1}{\sqrt{2^2+(-1)^2+5^2}\cdot \sqrt{1^2+(-3)^2+1^2} }=\dfrac{10}{\sqrt{30} \cdot \sqrt{11} }.$

$\cos\angle 3=\dfrac{\vec{b}\cdot (\vec{a}-\vec{b})}{|\vec{b}|\cdot |\vec{a}-\vec{b}|}=\dfrac{1\cdot 1+2\cdot (-3)+4\cdot 1}{\sqrt{1^2+2^2+4^2}\cdot \sqrt{1^2+(-3)^2+1^2} }=\dfrac{-1}{\sqrt{21} \cdot \sqrt{11} }.$

Then

$\angle 1=\arccos \left(\dfrac{20}{\sqrt{30} \cdot \sqrt{21} }\right)\approx 37.17^{\circ};$
$\angle 2=\arccos \left(\dfrac{10}{\sqrt{30} \cdot \sqrt{11} }\right)\approx 56.60^{\circ};$
$\angle 3=\arccos \left(\dfrac{-1}{\sqrt{21} \cdot \sqrt{11} }\right)\approx 93.77^{\circ}.$