3 years ago

X = y - 3

1 answer:

3 0

The answer is A. (1, 4), because when the values are substituted in to the equations, you get 1 = 4 - 3 and 1 + 12 = 13, which are both correct. I hope this helps!

You might be interested in

Is 25/99 a terminating decimal or a repeating decimal

BartSMP [9]

Repeating; it's .252525252525252525252525

3 0

3 years ago

Please help I need help

aliina [53]

It think the answer could quite possibly be 4

8 0

3 years ago

A stray dog ate 12 of your muffins. that was 3/10 of all of them! with how many did you start?

kolbaska11 [484]

Answer:

You started of with 84

4 0

3 years ago

I need help with this problem:)

mixer [17]

Answer:

11.5 for first question

4 0

3 years ago

Prove 2√(x) + 1/√(x + 1) <= 2√(x+1) for all x in [0,inf)

EleoNora [17]

Start by multiplying each side of the inequality by $\sqrt{x + 1}$ and simplifying:

$2 \sqrt x + \frac{1}{\sqrt{x+1}} \leq 2 \sqrt{x + 1}$
$(2 \sqrt x + \frac{1}{\sqrt{x+1}})(\sqrt{x + 1}) \leq (2 \sqrt{x + 1})(\sqrt{x + 1})$
$2 \sqrt{x(x + 1)} + 1 \leq 2(x + 1)$
$2 \sqrt{x^2 + x} + 1 \leq 2x + 2$
$2 \sqrt{x^2 + x} \leq 2x + 1$
$\sqrt{x^2 + x} \leq x + \frac{1}{2}$

From here, we can square both sides to get

$x^2 + x \leq (x + \frac{1}{2})^2$
$x^2 + x \leq x^2 + x + \frac{1}{4}$
$0 \leq \frac{1}{4}$ , which is always true.

3 0

3 years ago