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galina1969 [7]
3 years ago
14

The presentation of numerical information in a graph or table is called a data ___?

Mathematics
2 answers:
nlexa [21]3 years ago
8 0
Data graph, Data Chart or Bar graph in some cases.
professor190 [17]3 years ago
3 0
Its specificly called a 

Date Table Or Data Chart
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Use multiplication to find three ratios equivalent to 8:12
tatyana61 [14]
Ok so 8:12 can be 8 to 12, or 8\12

do that help?
4 0
3 years ago
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The length of s is 2/3 the length of t. If s has an area of 368 cm squared find the perimeter of the figure
ser-zykov [4K]
s = \frac{2}{3}t
\\\ \frac{2}{3}t*t = \frac{2}{3}t^2
\\\ \frac{2}{3}t^2 = 368
\\\ t^2 = \frac {3}{2}368
\\\ t^2 = 552
\\\ t \approx 23.5
\\\ s \approx 15.6
\\\ P \approx 78.2
3 0
3 years ago
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I will give u brainliest
spayn [35]

Answer:

9

Step-by-step explanation:

sqrt(85) is about 9.2, and the question asks to round to the nearest integer, making it 9. 9.2^2 is almost 85, which is how you can check your answer.

4 0
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.

8 0
3 years ago
Wayne used the diagram to compute the distance from Ferris, to Dunlap, to Butte. How much shorter is the distance directly from
OLEGan [10]

Answer:

10 mi

Step-by-step explanation:

In this question you have to use Pythagorus Theorem. a²+b²=c²

a = 15

b = 20

So, 15²+ 20² = c²

= 225 + 400 = 625

So Ferris to Butte is \sqrt{625} which is 25.

So now you would do 15 + 20 to find out Wayne's distance. 35 - 25 = 10.

Hope it made sense and I didn't waffle

7 0
3 years ago
Read 2 more answers
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