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3 years ago

15

Thanks an advance~! Show your work :)

1 answer:

Talja [164]3 years ago

6 0

Only AAS cannot prove this.

So the answer will be A.

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The point (4,-2) is the vertex of the graph of a quadratic function. The points (8,6) and (2,0) also fall on the graph of the fu

Elina [12.6K]

Answer:

Step-by-step explanation:

y = ax² + bx + c

~~~~~~~

(4, - 2), (8, 6), (2, 0)

a(4²) + b(4) + c = - 2

a(8²) + b(8) + c = 6

a(2²) + b(2) + c = 0

16a + 4b + c = - 2 .............. <em>(1)</em>

64a + 8b + c = 6 ............... <em>(2)</em>

4a + 2b + c = 0 ................. <em>(3)</em>

a = 0.5 ; b = - 4 ; c = 6

<em>y = 0.5x² - 4x + 6</em>

Part A: (0, 6), (6, 0)

Part B: (0,6)

Part C: (6, 0)

Part D: [ 4, ∞ )

4 0

3 years ago

Ok just answer this and im gonna mark as brainliest

suter [353]

Paperbacks and notebooks

3 0

3 years ago

Which statement correctly describes the graph of y = x + 9.5?

Vikki [24]

Answer:

A is the answer

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Find the coordinates of the centroid of triangle RST. SHOW YOUR WORK so I can see if the answer makes sense (no links)

zimovet [89]

Note the coordinates of each point: R(-4, 5), S(5, 1), T(2, -3).

The centroid is the point whose coordinates are the average of the coordinates of R, S, and T.

<em>x</em>-coordinate: (-4 + 5 + 2)/3 = 3/3 = 1

<em>y</em>-coordinate: (5 + 1 - 3)/3 = 3/3 = 1

So the centroid is (1, 1).

3 0

2 years ago

let x1,x2, and x3 be linearly independent vectors in R^(n) and let y1=x2+x1; y2=x3+x2; y3=x3+x1. are y1,y2,and y3 linearly indep

Nutka1998 [239]

Answer with Step-by-step explanation:

We are given that

$x_1,x_2$ and $x_3$ are linearly independent.

By definition of linear independent there exits three scalar $a_1,a_2$ and $a_3$ such that

$a_1x_1+a_2x_2+a_3x_3=0$

Where $a_1=a_2=a_3=0$

$y_1=x_2+x_1,y_2=x_3+x_2,y_3=x_3+x_1$

We have to prove that $y_1,y_2$ and $y_3$ are linearly independent.

Let $b_1,b_2$ and $b_3$ such that

$b_1y_1+b_2y_2+b_3y_3=0$

$b_1(x_2+x_1)+b_2(x_3+x_2)+b_3(x_3+x_1)=0$

$b_1x_2+b_1x_1+b_2x_3+b_2x_2+b_3x_3+b_3x_1=0$

$(b_1+b_3)x_1+(b_2+b_1)x_2+(b_2+b_3)x_3=0$

$b_1+b_3=0$

$b_1=-b_3$ ...(1)

$b_1+b_2=0$

$b_1=-b_2$ ..(2)

$b_2+b_3=0$

$b_2=-b_3$ ..(3)

Because $x_1,x_2$ and $x_3$ are linearly independent.

From equation (1) and (3)

$b_1=b_2$ ...(4)

Adding equation (2) and (4)

$2b_1==0$

$b_1=0$

From equation (1) and (2)

$b_3=0,b_2=0,b_3=0$

Hence, $y_1,y_2$ and $y_3$ area linearly independent.

5 0

2 years ago