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svetlana [45]
2 years ago
11

PLEASE HELP SHOW WORK

Mathematics
1 answer:
Pachacha [2.7K]2 years ago
4 0

Answer:

1) x = 18, y = -5

2) x= 1, -3  y = 3, -1

Step-by-step explanation:

1)

x+y=13

y = 13-x

2x-y=5

2x-(13-x) = 5

2x - 13 + x = 5

x - 13 = 5

x = 18

x + y = 13

18 + y = 13

y = -5

2)

y = x+2

x^2 + y^2 = 10

x^2 + (x+2)^2 = 10

x^2 + (x+2)(x+2) = 10

x^2 + x^2 + 4x + 4 = 10

2x^2 + 4x + 4 = 10

2x^2 + 4x - 6 = 0

2(x^2 + 2x - 3) = 0

2(x-1)(x+3) = 0

x = 1, x = -3

y = x+2

y = 1 +2 = 3

y = -3+2 = -1

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You like math then here
Igoryamba

Answer:

-8b + 9 - 5k

Step-by-step explanation:

I am assuming the blue highlighted portion is your answer, and is not a part of the initial question

(-4b + 15 - 7k) - (6 + 4b - 2k)

Combine like terms

-4b - + 4b

Negative + Positive = Negative

-4b - 4b = -8b

15 - 6 = 9

-7k - - 2k

Negative + Negative = Positive

-7k + 2k = -5k

Put them all together:

0b + 9 - 5k

Simplify

-8b + 9 - 5k

I got the same answer as you (just in a different order)

5 0
2 years ago
A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and
vredina [299]

Answer:

2.35%

Step-by-step explanation:

Mean number of months (M) = 39 months

Standard deviation (S) = 10 months

According to the 68-95-99.7 rule, 95% of the data is comprised within two standard deviations of the mean (39-20 to 39+20 months), while 99.7% of the data is comprised within two standard deviations of the mean (39-30 to 39+30 months).

Therefore, the percentage of cars still in service from 59 to 69 months is:

P_{59\ to\ 69}=\frac{P_{9\ to\ 69}-P_{19\ to\ 59}}{2} \\P_{59\ to\ 69}=\frac{99.7-95}{2}\\P_{59\ to\ 69}=2.35\%

The approximate percentage of cars that remain in service between 59 and 69 months is 2.35%.

8 0
2 years ago
-
posledela

Answer:

The bakery will bake 600 butterscotch bread, 550 chocolate bread, and 1,000 coconut bread

Step-by-step explanation:

In order to answer the word problem question, we list the parameters as follows;

The number of bread types the bakery takes = 3 types of bread

The monthly cost of the bakery for baking the bread, C = RM 6,850

The number of bread loaves baked = 2,150

The cost of baking a loaf of butterscotch bread = Rm 2

The cost of baking a loaf of chocolate bread = Rm 3

The cost of baking a loaf of coconut bread = Rm 4

The sale price of a butterscotch bread = Rm 3

The sale price of a chocolate bread = Rm 4.50

The sale price of a coconut bread = Rm 5

The amount of monthly profit the bakery makes, P = Rm 2,975

Therefore, the total revenue of the company, R = C + P

∴ R = Rm 6,850 + Rm 2,975 = Rm 9,825

Let 'x', 'y', and 'z', represent the number of loaves of butterscotch, chocolate, and coconut bread the company bakes respectively

Then we get;

x + y + z = 2,150...(1)

2·x + 3·y + 4·z = 6,850...(2)

3·x + 4.5·y + 5·z = 9,825...(3)

Multiplying equation (1) by 2 and subtracting from equation (2) gives;

2·x + 3·y + 4·z - 2×(x + y + z) = 6,850 - 2 × 2,150 = 2,550

2·x - 2·x + 3·y - 2·y + 4·z - 2·z = 2,550

∴ y + 2·z = 2,550...(4)

Adding equation (1) to (2) and subtracting from (3) gives;

3·x - 3·x + 4.5·y - 4·y + 5·z - 5·z = 9,825 - (6,850 + 2,150) = 825

1.5·y = 825

y = 825/1.5 = 550

The number of loaves of chocolate bread baked, y = 550

From equation (4), we get;

550 + 2·z = 2,550

2·z = 2,550 - 550 = 2,000

z = 2,000/2 = 1,000

The number of loaves of coconut bread baked, z = 1,000

∴ From equation (1) x = 2,150 - (550 + 1,000) = 600

The number of loaves of butterscotch bread baked, x = 600.

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