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alex41 [277]
3 years ago
10

Don’t understand #11 A,B, and C help please

Mathematics
1 answer:
Fiesta28 [93]3 years ago
4 0
I set it in a big problem. Since you know that all the angles of a triable add up to 180,
m<a + m<b + m<c = 180, plug in equations/values
(36) + (3x+12) + (3x+18) = 180, subtract 36
3x+12 + 3x+18 = 144, combine like terms,
6x+30 = 144, subtract 30,
6x=114, divide by 6,
x=19. Plug in X to the equations for m<b and m<c
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Can somebody help me please ?? :) Thank u if u do
Lorico [155]
SF = 9/18 = 1/2 or 0.5

answer
1/2 or 0.5
5 0
3 years ago
Deluxe River Cruises operates a fleet of river vessels. The fleet has two types of vessels: A type-A vessel has 60 deluxe cabins
bixtya [17]

Answer: It should be used 2 for type-A and 3 for type-B to minimize the cost.

Step-by-step explanation: As it is stipulated, <u>x</u> relates to type-A and y to type-B.

Type-A has 60 deluxe cabins and B has 80. It is needed a minimum of 360 deluxe cabins, so:

60x + 80y ≤ 360

For the standard cabin, there are in A 160 and in B 120. The need is for 680, so:

160x + 120y ≤ 680

To calculate how many of each type you need:

60x + 80y ≤ 360

160x + 120y ≤ 680

Isolating x from the first equation:

x = \frac{360 - 80y}{60}

Substituing x into the second equation:

160(\frac{360 - 80y}{60}) + 120y = 680

-3200y+1800y = 10200 - 14400

1400y = 4200

y = 3

With y, find x:

x = \frac{360 - 80y}{60}

x = \frac{360 - 80.3}{60}

x = 2

To determine the cost:

cost = 42,000x + 51,000y

cost = 42000.2 + 51000.3

cost = 161400

To keep it in a minimun cost, it is needed 2 vessels of Type-A and 3 vessels of Type-B, to a cost of $161400

8 0
2 years ago
Fraction of the diagram is shade
WITCHER [35]

Answer:

10/30 is the fraction of the diagram shade

8 0
3 years ago
What is the measure of this angle?
meriva

Answer: D. 170

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Solve the following system of equations: −2x + y = 1 −4x + y = −1
Katyanochek1 [597]

Answer:

<h2>x = 1, y = 3 → (1, 3)</h2>

Step-by-step explanation:

\left\{\begin{array}{ccc}-2x+y=1\\-4x+y=-1&\text{change the signs}\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}-2x+y=1\\4x-y=1\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad2x=2\qquad\text{divide both sides by 2}\\.\qquad x=1\\\\\text{Put it to the first equation:}\\\\-2(1)+y=1\\-2+y=1\qquad\text{add 2 to both sides}\\y=3

5 0
3 years ago
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