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oksian1 [2.3K]
3 years ago
11

If one redskin and 4 golden roughs cost $1.65, whereas 2 redskins and 3 golden roughs cost $1.55, how much does each type of swe

et cost?
Mathematics
1 answer:
pishuonlain [190]3 years ago
3 0

Answer:

redskin $0.50      golden rough $0.35

Step-by-step explanation:

r=redskin

g=golden rough

1r + 4g = 1.65

2r + 3g = 1.55

I will multiply the first by -2

-2r - 8g  = -3.30

2r  +3g  = 1.55

Add both

0 -5g = - 1.75

g= 0.35       0.35*4 = 1.4

r+ 1.4 = 1.65

r=0.25

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Subtract. Write the difference in simplest form.<br><br>(5n - 3) - (-2n + 7)​
kolezko [41]

Answer:

Simplifying

(5n + -3) + -1(-2n + 7) = 0

Reorder the terms:

(-3 + 5n) + -1(-2n + 7) = 0

Remove parenthesis around (-3 + 5n)

-3 + 5n + -1(-2n + 7) = 0

Reorder the terms:

-3 + 5n + -1(7 + -2n) = 0

-3 + 5n + (7 * -1 + -2n * -1) = 0

-3 + 5n + (-7 + 2n) = 0

Reorder the terms:

-3 + -7 + 5n + 2n = 0

Combine like terms: -3 + -7 = -10

-10 + 5n + 2n = 0

Combine like terms: 5n + 2n = 7n

-10 + 7n = 0

Solving

-10 + 7n = 0

Solving for variable 'n'.

Move all terms containing n to the left, all other terms to the right.

Add '10' to each side of the equation.

-10 + 10 + 7n = 0 + 10

Combine like terms: -10 + 10 = 0

0 + 7n = 0 + 10

7n = 0 + 10

Combine like terms: 0 + 10 = 10

7n = 10

Divide each side by '7'.

n = 1.428571429

Simplifying

n = 1.428571429

3 0
3 years ago
The expression 14p - 28 factored using the GCF is
Burka [1]

the factors and prime factorization of 14 and 28. (The biggest common factor number is the GCF number) So the greatest common factor 14 and 28 is 14.Jul

5 0
2 years ago
Two airplanes leave an airport at the same time, the first headed due north and the second at a bearing of N42^ E . At 2:00PM, t
Leno4ka [110]

Answer:

The two airplanes are about 330miles apart.

Step-by-step explanation:

The diagram interpreting the question has been attached to this response.

As shown in the diagram,

i. the airplanes leave at point C.

ii. at 2.00pm the first and second airplanes are at points A and B respectively, where they are 312miles and 487miles away from the starting point C in directions due north and N42E from the point C.

iii. the points A, B and C form a triangle with sides a, b and c.

To solve for the value of c which is the distance between the two planes at 2.00pm, the cosine rule is used.

c² = a² + b² - 2abcosC            --------------(i)

where;

b  = 312miles

a = 487miles

C = 42°

Substitute these values into equation (i) and solve as follows;

c² = (487)² + (312)² - 2(312)(487)cos(42)

c² = (237169) + (97344) - 303888cos(42)

c² = (237169) + (97344) - 303888(0.7431)

c² = 334513 - 225819.1728

c² = 108693.8272

<em>Take the square root of both sides</em>

√c² = √108693.8272

c = 329.69

c ≅ 330miles

Therefore, the two airplanes are far apart by 330miles

3 0
3 years ago
Use a property to explain why the product of a multiple of 10 and a number will have a zero in the ones place
Goryan [66]
The propriety is zero propriety. Because 0 times a number is 0
6 0
3 years ago
Please I need help with this and show your steps
krek1111 [17]
Here you go! Hope this helps!

3 0
2 years ago
Read 2 more answers
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