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3 years ago

compare the method of using expanded form and the method of using place value to multiply a decimal and a whole number

1 answer:

3 0

Unfortunately, multiplying and dividing decimals this effect does not occur. In some cases, the decimal number even complicates the operatsii.Dlya beginning, we introduce a new definition. We will meet him very often, and not only in this part of uroke.Znachaschaya - is all that is between the first and the last non-zero digit, including the ends. It's only about numbers, the decimal point is not uchityvaetsya.Tsifry included in the meaningful part of the number, are called significant figures. They may be repeated or even be zero.

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3 years ago

Triangle ABC has vertices A(-5, -2), B(7, -5), and C(3, 1). Find the coordinates of the intersection of the three altitudes

Darina [25.2K]

Answer:

Orthocentre (intersection of altitudes) is at (37/10, 19/5)

Step-by-step explanation:

Given three vertices of a triangle

A(-5, -2)

B(7, -5)

C(3, 1)

Solution A by geometry

Slope AB = (yb-ya) / (xb-xa) = (-5-(-2)) / (7-(-5)) = -3/12 = -1/4

Slope of line normal to AB, nab = -1/(-1/4) = 4

Altitude of AB = line through C normal to AB

(y-yc) = nab(x-xc)

y-1 = (4)(x-3)

y = 4x-11 .........................(1)

Slope BC = (yc-yb) / (xc-yb) = (1-(-5) / (3-7)= 6 / (-4) = -3/2

Slope of line normal to BC, nbc = -1 / (-3/2) = 2/3

Altitude of BC

(y-ya) = nbc(x-xa)

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

y = 2x/3 + 10/3 - 2

y = (2/3)(x+2) ........................(2)

Orthocentre is at the intersection of (1) & (2)

Equate right-hand sides

4x-11 = (2/3)(x+2)

Cross multiply and simplify

12x-33 = 2x+4

10x = 37

x = 37/10 ...................(3)

substitute (3) in (2)

y = (2/3)(37/10+2)

y=(2/3)(57/10)

y = 19/5 ......................(4)

Therefore the orthocentre is at (37/10, 19/5)

Alternative Solution B using vectors

Let the position vectors of the vertices represented by

a = <-5, -2>

b = <7, -5>

c = <3, 1>

and the position vector of the orthocentre, to be found

d = <x,y>

the line perpendicular to BC through A

(a-d).(b-c) = 0 "." is the dot product

expanding

<-5-x,-2-y>.<4,-6> = 0

simplifying

6y-4x-8 = 0 ...................(5)

Similarly, line perpendicular to CA through B

<b-d>.<c-a> = 0

<7-x,-5-y>.<8,3> = 0

Expand and simplify

-3y-8x+41 = 0 ..............(6)

Solve for x, (5) + 2(6)

-20x + 74 = 0

x = 37/10 .............(7)

Substitute (7) in (6)

-3y - 8(37/10) + 41 =0

3y = 114/10

y = 19/5 .............(8)

So orthocentre is at (37/10, 19/5) as in part A.

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3 years ago

What is a turning point on a quadratic graph?

Cerrena [4.2K]

Answer:

A turning point is the highest or lowest point on a quadratic graph.

Step-by-step explanation:

A quadratic graph looks something like the graph below.

The equation of a quadratic graph would normally look like

+/- ax^2 + bx + c

An example might be -16x^2 + 5x + 4

Note the negative symbol in front of the 16. The negative means that the graph will be facing downwards, or that the turning point is the highest point. A positive graph will mean that the graph is facing upwards, or that the turning point is the lowest point.

Essentially, it is the location where a graph has its lowest or highest point and where the y-values (can include x-values in horizontal quadratics) "turn" to the direction they originated.

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2 years ago