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3 years ago

Solve each system using the substitution method 6x-y=-4 and 2x+2y=15

1 answer:

8 0

Convert 6x - y = -4 into y = mx + b
Y - 4 = 6x
Y = 6x + 4
Substitute that into second equation
2x + 2(6x + 4) = 15
2x + 12x + 8 = 15
14x = 7
X = 1/2
6(1/2) - y = -4
3 - y = -4
-y = -7, y = 7
Final answer: x = 1/2, y = 7

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If you help with the correct answer u are a god .pt2

Brrunno [24]

a. Answer: D: (∞, ∞)

R: (-∞, ∞)

Step-by-step explanation:

Theoretical domain is the domain of the equation (without an understanding of what the x-variable represents).

Theoretical range is the range of the equation given the domain.

c(p) = 25p

There are no restrictions on the p so the theoretical domain is All Real Numbers.

Multiplying 25 by All Real Numbers results in the range being All Real Numbers.

a) D: (∞, ∞)

R: (-∞, ∞)

*********************************************************************************

b. Answer: D: (0, 200)

R: (0, 5000)

Step-by-step explanation:

Practical domain is the domain of the equation WITH an understanding of what the x-variable represents.

Practical range is the range of the equation given the practical values of the domain.

The problem states that p represents the number of cups. Since we can't have a negative amount of cups, p ≥ 0. The problem also states that Bonnie will purchase a maximum of 200 cups. So, 0 ≤ p ≤ 200

The range is 25p → (25)0 ≤ (25)p ≤ (25)200

→ 0 ≤ 25p ≤ 5000

b) D: (0, 200)

R: (0, 5000)

5 0

3 years ago

8. Is the following statement true or false? Justify your conclusion. If a and b are nonnegative real numbers and a + b =0, then

atroni [7]

Answer:

If $a + b = 0$ then $a = 0$ and $b =0$

a | b | a + b (answer)

0 | 0 | 0

0 | 1 | 1

0 | 2 | 2

1 | 0 | 1

2 | 0 | 2

1 | 1 | 2

2 | 1 | 3

Step-by-step explanation:

Considering the following conditions for the real numbers:

$a\geq 0\\b\geq 0$

Following the rules of these in-equations, it is possible to deduce:

$a + b \geq 0$

Then, if the proposed statement is:

$a + b = 0$

The conditions above shall comply the requirements established, but first, analyzing the statement:

If $a \geq 0$ and $b \geq 0$ then $a + b \geq a$ , $a + b \geq b$ and $a + b \geq 0$ .

If $a = 0$ and b a non negative real number, then $a + b \geq b$ , but because to $a = 0$ , then $a + b = b$ . Due to the commutative property of sums, the same behavior will be presented if $b = 0$ and a a non negative real number.