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Westkost [7]
3 years ago
9

A fish dives into a depth of -3/4 mile below the surface of the water. A turtle's depth is 1/3 as deep as the fish's depth. What

is the turtle's depth?
Mathematics
1 answer:
Ksenya-84 [330]3 years ago
8 0
For this case, what we must do is multiply the depth of the fish by the fraction of depth of the turtle.
 We have then:
 (-3/4) * (1/3) = (- 3 * 1) / (4 * 3) = - 1/4 mile
 Therefore, the depth of the turtle on the surface of the water is:
 -1/4 mile
 Answer:
 the turtle's depth is:
 -1/4 mile
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What is that product of 2x + y and 5x -y + 3
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Answer:

10x² + 6x + 3xy + 3y - y²

Step-by-step explanation:

Each term in the second factor is multiplied by each term in the first factor, as shown

(2x + y)(5x - y + 3)

= 2x(5x - y + 3) + y(5x - y + 3) ← distributing

= 10x² - 2xy + 6x + 5xy - y² + 3y ( collect like terms )

= 10x² + 6x + 3xy + 3y - y²


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Reducing fraction 80/100
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4/5 is fully reduced


Step-by-step explanation:


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Libby spent 2.50 on 10 picies of gum at this rate how much will 22 piecies of gum cost
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$5.50

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6 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
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