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Murljashka [212]
3 years ago
8

given triangle abc is similar to triangle efg, with medians cd and gh respectively, ac=9, eg=6, cd=x+2, gh=2x-4. find the length

of cd
Mathematics
1 answer:
Leno4ka [110]3 years ago
3 0

Answer:

X= 9//

Step-by-step explanation:

abc = efg

ac + cd = eg + gh

9 + (x+2) = 6 + (2x-4)

9 - 6 = 2x-4 - (x+2)

3 = x - 6

3+6 = x

9 = x//

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Yes.

Step-by-step explanation:

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Convert degree into Radian that is 18 degree 12'​
kiruha [24]

Answer:

\frac{\pi }{10} radians

Step-by-step explanation:

To convert degrees into radians, multiply the degrees with \frac{\pi}{180}.

If you have 18 degrees:

18*\frac{\pi}{180} = \frac{\pi }{10}

4 0
3 years ago
James has eight coins in his pocket. All the coins are either dimes or quarters.The total value of the coins is $1.25. How many
Goshia [24]

Answer:

he has 5d+3q

Step-by-step explanation:

since every d is .10 and every q is .25 the equation can be changed to 5(.10)+3(.25) which is equal to .5+.75 which adds up to 1.25

i did this by assuming that the amount of d is 8 at first, then change one of the 8 d to a q which will give you 7d+q, then assume another d is a q which will give you 6d+2q, repeat this until the amount of d and q add up to 1.25

6 0
3 years ago
Southern Oil Company produces two grades of gasoline: regular and premium. The profit contributions are $0.30 per gallon for reg
Contact [7]

Answer:

a) MAX--> PC (R,P) = 0,3R+ 0,5P

b) <u>Optimal solution</u>: 40.000 units of R and 10.000 of PC = $17.000

c) <u>Slack variables</u>: S3=1000, is the unattended demand of P, the others are 0, that means the restrictions are at the limit.

d) <u>Binding Constaints</u>:

1. 0.3 R+0.6 P ≤ 18.000

2. R+P ≤ 50.000

3. P ≤ 20.000

4. R ≥ 0

5. P ≥ 0

Step-by-step explanation:

I will solve it using the graphic method:

First, we have to define the variables:

R : Regular Gasoline

P: Premium Gasoline

We also call:

PC: Profit contributions

A: Grade A crude oil

• R--> PC: $0,3 --> 0,3 A

• P--> PC: $0,5 --> 0,6 A

So the ecuation to maximize is:

MAX--> PC (R,P) = 0,3R+ 0,5P

The restrictions would be:

1. 18.000 A availabe (R=0,3 A ; P 0,6 A)

2. 50.000 capacity

3. Demand of P: No more than 20.000

4. Both P and R 0 or more.

Translated to formulas:

Answer d)

1. 0.3 R+0.6 P ≤ 18.000

2. R+P ≤ 50.000

3. P ≤ 20.000

4. R ≥ 0

5. P ≥ 0

To know the optimal solution it is better to graph all the restrictions, once you have the graphic, the theory says that the solution is on one of the vertices.

So we define the vertices: (you can see on the graphic, or calculate them with the intersection of the ecuations)

V:(R;P)

• V1: (0;0)

• V2: (0; 20.000)

• V3: (20.000;20.000)

• V4: (40.000; 10.000)

• V5:(50.000;0)

We check each one in the profit ecuation:

MAX--> PC (R,P) = 0,3R+ 0,5P

• V1: 0

• V2: 10.000

• V3: 16.000

• V4: 17.000

• V5: 15.000

As we can see, the optimal solution is  

V4: 40.000 units of regular and 10.000 of premium.

To have the slack variables you have to check in each restriction how much you have to add (or substract) to get to de exact (=) result.  

3 0
3 years ago
Tan 2θ; cos θ = 8 17 , θ in Quadrant I
SOVA2 [1]
The answer would be (20, 8) because it is in quadrant 1 and all numbers are posiive, so all you have to do is find wich one is the x codinate and wich one is y
5 0
3 years ago
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