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Sholpan [36]
3 years ago
9

Joe bought some CD's and DVD's at the store.

Mathematics
1 answer:
Elina [12.6K]3 years ago
7 0

Answer: x=8 and y=4

Step-by-step explanation:

x+y=12           x=12-y                     x=12-4

14x+22y=200                                x=8

14(12-y)+22y=200

168-14y+22y=200

-168                 -168

-14y+22y=32

8y=32

8y/8=32/8

y=4

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1.Write the expression in expanded form. The subscript number can be write in
enyata [817]

Answer:

(a) Expanded form:

((X_1 + X_1Y_1) - X_1Z_1) + ((X_2 + X_2Y_2) - X_2Z_2) + ((X_3 + X_3Y_3) - X_3Z_3) + ((X_4 + X_4Y_4) - X_4Z_4)

(b) The value of the expression: -21

Step-by-step explanation:

Given

\sum \limit^4_{i=1}\ ((X_i+X_iY_i) - X_iZ_i)

Solving (a): The expanded form:

This means that we substitute the values of i from 1 to 4 in the above expression.

So, the expression becomes:

((X_1 + X_1Y_1) - X_1Z_1) + ((X_2 + X_2Y_2) - X_2Z_2) + ((X_3 + X_3Y_3) - X_3Z_3) + ((X_4 + X_4Y_4) - X_4Z_4)

Solving (b): The value of the expression

To do this, we simply substitute the given values of X1, X2....... in the expression.

This gives:

So, the expression becomes:

((0 + 0*5) - 0*0) + ((1 + 1*26) - 1*-2) + ((12 + 12*-2) - 12*3) + ((-1 + -1*25) - -1*24)

Simplify each bracket

((0 + 0) - 0) + ((1 + 26) +2) + ((12 -24) - 36) + ((-1 -25) +24)

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<em>Hence, the result of the expression is -21</em>

5 0
3 years ago
Find two consecutive even integers such that the square of the second, decreased by twice the first is 52
Fiesta28 [93]

Let the first integer be x. So, the next even number is x+2

Let's translate the formula: the square of the second is (x+2)^2, twice the first is 2x. So, the equation is

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Expand the square to get

x^2+4x+4-2x = 52 \iff x^2+2x+4 = 52 \iff x^2+2x - 48 = 0

The solutions to this equations are

x = -8,\quad x = 6

So, the possible consecutive even numbers are

-8\text{ and } -6\quad\text{or}\quad 6\text{ and } 8

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tia_tia [17]
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Step-by-step explanation:

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you can also switch the numbers your answer will always be the same

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