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chubhunter [2.5K]
3 years ago
10

An expression is shown below:

Mathematics
2 answers:
olganol [36]3 years ago
7 0

Answer:

An expression is shown below:

10n³− 15n² + 20xn² − 30xn

Part A: Rewrite the expression so the GCF is factored completely. (4 points)

10n³− 15n² + 20xn² − 30xn

2*5*n*n*n-5*3*n*n+2*5*2*x*n*n-2*5*3*x*n

<u>Greatest</u><u> </u><u>common</u><u> </u><u>factor</u><u>=</u><u>5</u><u>*</u><u>n</u><u>=</u><u>5</u><u>n</u>

Part B: Rewrite the expression completely factored. Show the steps of your work.

Solution given;

10n³− 15n² + 20xn² − 30xn

5n(2n²-3n+4xn-6x)

5n(2n²+4xn-3n-6x)

5n(2n(n+2x)-3(n+2x))

<u>5n(n+2x)(2n-3</u>)

lilavasa [31]3 years ago
4 0

Answer:

5n(2n-3)(n+2x)

Step-by-step explanation:

<u>Step 1:   Rewrite the expression so the GCF is factored completely</u>

<u />10n^{3} - 15n^{2} + 20xn^{2} - 30xn

The GCF is 5n so factor it out

5n(2n^{2} - 3n + 4xn - 6x)

<u>Step 2:  Rewrite the expression completely factored</u>

<u />5n(2n^{2} - 3n + 4xn - 6x)

5n(2n(n+2x)-3(n+2x))

5n(2n-3)(n+2x)

Answer:  5n(2n-3)(n+2x)

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find the equation of the sides of an isosceles right angled triangle whose vertex is (-2,-3) and the base is on the line x=0​
Serggg [28]

Answer:

AC:y=x-1  CB:y=-x-5 AB:x=0

Step-by-step explanation:

Consider the triangle. The base AB is on the line x=0, the vertex C is (-2,-3)

The side AC is equal to BC. The angle ACB is 90 degrees. If the base is on the line x=o, it is on the axis Y.Explore the distance from the point C to the AB

c(-2,-3), the distance to the axis Y is equal to the modul of the coordinate x (-2), it is 2. The coordinates of point projected by the point C to the axis Y is N(0,-3). The modul of the height is 2, the height of the isosceles triangle to the base is the bisectrix, so the angle BCA is 90/2=45degrees, CBA is 180-90-45=45 degrees too

the heigt CN is equal to side NB, NB=2

Suppose B is (0,y) (x=0 because the base is on this line)

THe modul of the vector NB is equal to sqrt ((0-0)^2+(y+3)^2)= 2

modul (y+3)= 2

y=-1 or y=-5

(0,-1), (0,-5) - two points, one of them (suppose B) is (0,-5) when A is (0,-1) (A is remote from the point N on the same distance with B, because AB is the median too)

Find CB and AC

Use the equation for AC

(x-0)/(-2-0)= (y+1)/(-3+1)

x/-2= (y+1)/-2

x=y+1

y=x-1

For CB

(x-0)/ (-2-0)= (y+5)/ (-3-(-5))

x/-2= (y+5)/2

-x=y+5

y=-x-5

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