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aleksandrvk [35]
3 years ago
6

Which of the following statement is correct​

Mathematics
1 answer:
bagirrra123 [75]3 years ago
5 0

Answer:

the 3rd one

Step-by-step explanation:

mark me brainliest if it was helpful

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What is an equation for the linear function whose graph contains the points (−1, −2) and (3, 10) ? Enter your answers in the box
balandron [24]
\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~{{ -1}} &,&{{ -2}}~) 
%  (c,d)
&&(~{{ 3}} &,&{{ 10}}~)
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}\implies 
\cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{10-(-2)}{3-(-1)}\implies \cfrac{10+2}{3+1}
\\\\\\
\cfrac{12}{4}\implies \cfrac{3}{1}\implies 3

\bf \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies 
y-(-2)=3[x-(-1)]
\\\\\\
y+2=3(x+1)
\implies 
y+2=3x+3\implies y=3x+1
4 0
3 years ago
Read 2 more answers
Can someone answer the question below?
devlian [24]

the answer should be m= -13

5 0
3 years ago
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4. Jenny works at a package delivery store.
Contact [7]
I think the answer is 525
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4 years ago
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How do you round 1866.025404 to the nearest centimetre? Please explain.
Vera_Pavlovna [14]

Answer:

1866

Step-by-step explanation:

1866 because the next digit is zero , therefore the value to the nearest whole number is 1866.

8 0
3 years ago
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
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