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gogolik [260]
3 years ago
10

What is the solution to the system of equations 6x+2y=6 7x+3y=5

Mathematics
1 answer:
brilliants [131]3 years ago
8 0

Answer:

(x,y)=(2,-3)

Step-by-step explanation:

6x+2y=6

7x+3y=5

Multiply both sides by 3

18x+6y=18

7x+3y=5

Multiply both sides by -2

18x+6y=18

-14x-6y=-10

Eliminate one variable by adding the equations

4x=8

x=2

Substitute the value of x to get y

6(2)+2y=6

y=-3

x=2

y=-3

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130°

Step-by-step explanation:

a straight line is 180°

so you have

180°-50°= 130°

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You’ve a sponge that is 66 square inches. How many times would you use 3 inches and 1 inch length
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Alexandra has some dimes and some quarters. She has no more than 21 coins worth a minimum of $3.75 combined. If Alexandra has 4
olga55 [171]

Answer:

A minimum of 10 dimes and 11 quarters is what Alexandra will have

Step-by-step explanation:

Let

d = number of dimes

q = number of quarters

Since she has 21 coins altogether,

d + q = 21------------------------equation 1    

  •  If these coins are worth $3.75 then

0.10 x d + 0.25 x q = 3.75

  • which is 0.10d +.25q =3.75 --------------------------equation 2

where $.10 is the value of one dime and $.25 is the value of one quarter

make d the subject of formula from equation 1 d = 21 -q----------equation 3

 insert it in equation 2

0.10d +0.25q =3.75

0.10(21-q) + 0.25q = 3.75

0.1(21)-0.1q+0.25q=3.75

2.1 +0.15q = 3.75

0.15q  = 3.75-2.1 = 1.65

q = 1.65/0.15 =165/15 =11

  • since we have the value of q insert in equation 3

d = 21 - q

d = 21-11

d = 10

Alexandra has 10 dimes and 11 quarters.

from my calculation i can see that the a minimum of 10 dimes and 11 quarters is what Alexandra will have

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3 years ago
Evaluate this expression if r = 2, t = -7, and u = -5<br> 3(r+u)
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4. From a faculty of six professors, six associate professors, ten assistant professors, and twelve instructors, a committee of
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Answer:

P=0.228

Step-by-step explanation:

We know that from a faculty of six professors, six associate professors, ten assistant professors, and twelve instructors, a committee of size six is to be selected.

Therefore, we have 34 people.

We calculate the number of possible combinations:

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Of the 6 professors we choose 2, and of the other 28 people we choose 4.

We calculate the number of favorable combinations:

C_2^6\cdot C_4^{28}=\frac{6!}{2!(6-2)!}\cdot \frac{28!}{4!(28-4)!}=15\cdot 20475=307125\\

Therefore, the probability is:

P=\frac{307125}{1344904}=0.228

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3 years ago
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