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£60 is divided between Peter, Angad & Nick so that Peter gets twice as much as Angad, and Angad gets three times as much as

Anna [14]

60£

Peter = 2 Angard

Angard = 3 Nick

Peter = 2 Angard = 6 Nick

Angard = 3 Nick

Nick = 1 Nick

6+3+1=10 Nick

60£/10 = 6£

Nick = 6£

Angvard = 3 Nick = 6£*(3) = 18£

Peter = 6 Nick = 6£*(6) = 36£

(prove: 6+18+36=60£)

3 0

3 years ago

A family has two cars. the first car has a fuel efficiency of 30 miles per gallon of gas and the second has a fuel efficiency of

jeka57 [31]

X : the gas used by first car in 1 particular week
y : the gas used by second car in 1 particular week
$x + y = 60 \\ x = 60 - y$
$30x + 35y = 1975 \\ 30 \times (60 - y) + 35y = 1975 \\ ... \\ y = 35 \\ x = 60 - y = 60 - 35 = 25$

5 0

3 years ago

ALGEBRAIC EXPRESSION 11. Subtract the sum of 13x – 4y + 7z and – 6z + 6x + 3y from the sum of 6x – 4y – 4z and 2x + 4y – 7. 12.

Naily [24]

Answer:

Explained below.

Step-by-step explanation:

(11)

Subtract the sum of (13x - 4y + 7z) and (- 6z + 6x + 3y) from the sum of (6x - 4y - 4z) and (2x + 4y - 7z).

$[(6x - 4y - 4z) +(2x + 4y - 7z)]-[(13x - 4y + 7z) + (- 6z + 6x + 3y) ]\\=[6x-4y-4z+2x+4y-7z]-[13x-4y+7z-6z+6x+3y]\\=6x-4y-4z+2x+4y-7z-13x+4y-7z+6z-6x-3y\\=(6x+2x-13x-6x)+(4y-4y+4y-3y)-(4z+7z+7z-6z)\\=-11x+y-12z$

Thus, the final expression is (-11x + y - 12z).

(12)

From the sum of (x² + 3y² - 6xy), (2x² - y² + 8xy), (y² + 8) and (x² - 3xy) subtract (-3x² + 4y² - xy + x - y + 3).

$[(x^{2} + 3y^{2} - 6xy)+(2x^{2} - y^{2} + 8xy)+(y^{2} + 8)+(x^{2} - 3xy)] - [-3x^{2} + 4y^{2} - xy + x - y + 3]\\=[x^{2} + 3y^{2} - 6xy+2x^{2} - y^{2} + 8xy+y^{2} + 8+x^{2} - 3xy]- [-3x^{2} + 4y^{2} - xy + x - y + 3]\\=[4x^{2}+3y^{2}-xy+8]-[-3x^{2} + 4y^{2} - xy + x - y + 3]\\=4x^{2}+3y^{2}-xy+8+3x^{2}-4y^{2}+xy-x+y-3\\=7x^{2}-y^{2}-x+y+5$