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solong [7]
3 years ago
7

Create an addition or subtraction problem involving polynomial expressions so that the answer can be shown using polynomial tile

s. Use the Graphing Calculator to verify your solution. Save a snapshot for others to reference as they attempt your challenge.
Mathematics
1 answer:
VLD [36.1K]3 years ago
7 0

Answer:

a^2 - b^2

Step-by-step explanation:

a ( a - b ) + b ( a - b (

= a^2 - ab + ba - b^2

= a^2 - ab + ab - b^2

= a^2 + 0 - b^2

= a^2 - b^2

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I need some help on this lol
nasty-shy [4]
I think it’s option A
3 0
3 years ago
How do you find the area of a polygon in a coordinate plane
inna [77]

Step-by-step explanation:

The way that I would do it, would be to use the point distance formula to find the length of each side.

Distance formula: \sqrt{(x2-x1)^{2} (y2-y1) ^{2} }

After you use this formula to find the length of all sides of the polygon, you would need to use the appropriate area formula to find the area of the polygon.

Area of Triangle: A= 1/2 base times height

Area of Rectangle: A= length times width

Area of Square: A= s^2 where s is the length of the side

Area of Parallelogram: A= base times height

Area of Trapezoid: A= 1/2 times height (base 1 + base 2) where base 1 and base 2 are the parallel sides

Area of Kite or Rhombus: A= 1/2 (diagonal 1 times diagonal 2)

Area of Regular Polygons: A= 1/2 (apothem times perimeter)

6 0
3 years ago
If X= -6 and y=4 what is x² + y²​
Rainbow [258]

Answer:

52

Step-by-step explanation:

(-6)x(-6)=36

(4)x(4)=16

36+16=52

6 0
2 years ago
Read 2 more answers
Answer as soon as possible
icang [17]

Answer:

  B.  9, 9x, 18x

Step-by-step explanation:

The value in each box is the product of the row heading and column heading. You can find the missing column heading by dividing the box value (162) by the row heading (18).

7 0
4 years ago
Read 2 more answers
Cos(a) 63/65 to find sin(a) and tan(a)
Semenov [28]

Answer:

\huge\boxed{\sin\alpha=-\dfrac{16}{65},\ \tan\alpha=-\dfrac{16}{63}}\\\\or\\\\\huge\boxed{\sin\alpha=\dfrac{16}{65},\ \tan\alpha=\dfrac{16}{63}}

Step-by-step explanation:

\cos\alpha=\dfrac{63}{65}\\\\\text{use}\ \sin^2\alpha+\cos^2\alpha=1\\\\\sin^2\alpha+\left(\dfrac{63}{65}\right)^2=1\\\\\sin^2\alpha+\dfrac{3969}{4225}=1\qquad\text{subtract}\ \dfrac{3969}{4225}\ \text{from both sides}\\\\\sin^2\alpha=\dfrac{4225}{4225}-\dfrac{3969}{4225}\\\\\sin^2\alpha=\dfrac{256}{4225}\to\sin\alpha=\pm\sqrt{\dfrac{256}{4225}}\\\\\sin\alpha=\pm\dfrac{\sqrt{256}}{\sqrt{4225}}\\\\\sin\alpha=\pm\dfrac{16}{65}

\text{use}\ \tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}\\\\\text{substitute:}\\\\\tan\alpha=\dfrac{\pm\frac{16}{65}}{\frac{63}{65}}=\pm\dfrac{16}{65}\cdot\dfrac{65}{63}=\pm\dfrac{16}{63}

3 0
4 years ago
Read 2 more answers
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