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just olya [345]
3 years ago
13

Hours worked / pay

Mathematics
1 answer:
Nadusha1986 [10]3 years ago
7 0
It is c because 8x2=16 and 8x4=32 and so on
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A store purchases a bicycle from the manufacturer for $150. The store then sells the bicycle for $215. By what percentage did th
Nookie1986 [14]

Answer:

it marked the bike up by 69 percent

Step-by-step explanation:

7 0
3 years ago
What is the missing number that makes the ratios equivalent? <br> 12:9<br> _:18​
S_A_V [24]

Answer:

24

Step-by-step explanation:

7 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
What is the solution to: <br> T/t-4=t+4/6
S_A_V [24]
\frac{t}{t-4}= \frac{t+4}{6} \\ t \neq 4 \\ 6t=  (t+4)(t-4) \\ 6t=t^2-16 \\ t^2-6t-16=0 \\ D=b^2-4ac=(-6)^2-4*1*(-16)=36+64=100 \\ t_{1,2}= \frac{-bб \sqrt{D} }{2a}  \\ t_1= \frac{6-10}{2}=-2 \\  \ t_2= \frac{6+10}{2}=8

Answer: t=-2, t=8
4 0
3 years ago
The circle with center (1,5) and radius 4 has equation
Gala2k [10]
<span>x2</span>+<span>y2</span>+6x+10y+9=<span>0
that is ur answer :)</span>
5 0
3 years ago
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