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Misha Larkins [42]
3 years ago
8

Solve for x -2.3(r - 1.2 ) = -9.66 enter you answer as decimal.

Mathematics
1 answer:
frozen [14]3 years ago
3 0

Answer: 5.4


Step-by-step explanation: Isolate the variable by dividing each side factors that dn't dontain the variable.

Hope this helps! :) ~Zane


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-22/5 +4/5

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Let $A = (4,-1)$, $B = (6,2)$, and $C = (-1,2)$. There exists a point $X$ and a constant $k$ such that for any point $P$,
goldfiish [28.3K]

Answer:

k=32

Step-by-step explanation:

Given the points:

A = (4,-1)$, $B = (6,2)$, and $C = (-1,2)$.

The first step is to find the <u>Centroid</u> of the triangle.

Centroid, X

=\left(\dfrac{x_1+x_2+x_3}{3} ,\dfrac{y_1+y_2+y_3}{3} \right)\\=\left(\dfrac{4+6+(-1)}{3} ,\dfrac{-1+2+2}{3} \right)\\=\left(\dfrac{9}{3} ,\dfrac{3}{3} \right)=(3,1)

Next, let P be a point (x,y)

Using the <u>distance formula, </u>\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}<u />

<u />PA^2=(x-4)^2+(y-(-1))^2\\PB^2=(x-6)^2+(y-2)^2\\PC^2=(x-(-1))^2+(y-2)^2\\PX^2=(x-3)^2+(y-1)^2\\<u />

On Substitution into: PA^2 + PB^2 + PC^2 = 3PX^2 + k

(x-4)^2+(y-(-1))^2+(x-6)^2+(y-2)^2+(x-(-1))^2+(y-2)^2=3[(x-3)^2+(y-1)^2]+k

Let us simplify the LHS first

\\LHS: x^2-8x+16+y^2+2y+1+x^2-12x+36+y^2\\-4y+4+x^2+2x+1+y^2-4y+4\\=3x^2-18x+3y^2-6y+62

Also, the Right Hand Side

RHS:3[(x-3)^2+(y-1)^2]+k\\=3[x^2-6x+9+y^2-2y+1]+k\\=3x^2-18x+27+3y^2-6y+3+k\\=3x^2+3y^2-18x-6y+30+k

Therefore:

3x^2-18x+3y^2-6y+62=3x^2+3y^2-18x-6y+30+k\\k=3x^2-18x+3y^2-6y+62-3x^2-3y^2+18x+6y-30\\k=3x^2-3x^2+3y^2-3y^2-18x+18x+62-30\\k=32

7 0
3 years ago
What is 20:50 (Equivalent Ratios)
Arturiano [62]

Answer: 20 : 50 (m1 = 1) 40 : 100 (m2 = 2) 60 : 150 (m3 = 3) 80 : 200 (m4 = 4) 100 : 250 (m5 = 5) . Bababoi

Hope I helped

Step-by-step explanation:

8 0
3 years ago
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