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3 years ago

HELP PLEASE I DONT UNDERSTAND

Mathematics

1 answer:

miv72 [106K]3 years ago

5 0

Answer:

None of these

Step-by-step explanation:

A perpendicular bisector is a segment which intersects a given segment at a 90° angle, and passes through the given segment's midpoint. Since point T divides segment SU into two parts with lengths 46 and 62 units, VT is not a perpendicular bisector.

A median of the triangle is a segment joining a vertex to the midpoint of the opposite side. Since point T divides segment SU into two parts with lengths 46 and 62 units, VT is not a median.

An altitude of the triangle is a segment passing through a vertex and perpendicular to (i.e., forming a right angle with) a line containing the base (the side opposite the vertex). The diagram doesn't show the right angle at point T (VT is not perpendicular to SU), then VT is not an altitude.

Thus, option None of These is true.

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Answer:

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Step-by-step explanation:

(10,11) means x=10,y=11

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maxonik [38]

Answer:

y = -3x - 5

Step-by-step explanation:

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4 years ago

Graph the functions on the same coordinate axis. {f(x)=−2x+1g(x)=x2−2x−3

NikAS [45]

Answer:

(2,-3) and (-2,5)

Step-by-step explanation:

Let us graph the two equations one by one.

1. $f(x)=-2x+1$

If we compare this equation with the slope intercept form of a line which is given as

$y=mx+c$

we see that m = -1 and c =1

Hence the slope of the line is -2 and the y intercept is 1. Hence one point through which it is passing is (0,1) .

Let us find another point by putting x = 1 and solving it for y

$y=-2(1)+1$

$y=-2+1 = -1$

Let us find another point by putting x = 2 and solving it for y

$y=-2(2)+1$

$y=-4+1 = -3$

Hence the another point will be (2,-3)

Let us find another point by putting x = -2 and solving it for y

$y=-2(-2)+1$

$y=+1 = 5$

Hence the another point will be (-2,5)

Now we have two points (0,1) ,(1,-1) , (2,-3) and (-2,5) we joint them on line to obtain our line

$g(x)=y=x^2-2x-3$

$y=x^2-2x+1-1-3$

$y=(x-1)^2-4$

$(y+4)=(x-1)^2$

It represents the parabola opening upward with vertices (1,-4)

Let us mark few coordinates so that we may graph the parabola.

i) x=0 ; $y=y=(0)^2-2(0)-3=0-0-3=-3$ ; (0,-3)

ii)x=-1 ; $y=(-1)^2-2(-1)-3=1+2-3=0$ ; (-1,0)

iii) x=2 ; $y=(2)^2-2(2)-3 = 4-4-3 =-3$ ;(2,-3)

iii) x=1 ; $y=(1)^2-2(1)-3 = 1-2-3 =-4$ ;(1,-4)

iii) x=-2 ; $y=(-2)^2-2(-2)-3 = 4+4-3 =5$ ;(-2,5)

Now we plot them on coordinate axis and line them to form our parabola

When we plot them we see that we have two coordinates (2,-3) and (-2,5) are common , on which our graphs are intersecting. These coordinates are solution to the two graphs.

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4 years ago