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3 years ago

Solve the system of equations - 3x + y = 7 and 9x – 6y = 3 by combining the equations.

Mathematics

1 answer:

aniked [119]3 years ago

3 0

(-5, -8)
(x, y)
would be your answer

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Marco has $38.43 dollars in his checking account. His checking account is linked to his amazon music account so he can buy music

Mumz [18]

Step-by-step explanation:

38.43 - 1.29x

Variable x represents the number of songs purchased

Not sure why this question didn't save the answer I just wrote (I'm new to the app)

5 0

3 years ago

Chanelle deposits $7,500 into the bank. She does not withdraw or deposit money for 6 years. She earns 6% interest during that ti

koban [17]

Answer: a. $2700

b. $10200

Step-by-step explanation:

a. The interest she would have earned at the end of the 6 years can be gotten using the formula

= PRT/100

= $7500 × 6% × 6

= $7500 × 6/100 × 6

= $7500 × 0.06 × 6

= $2700

b. Her balance when she wants to withdraw the money would be:

= $7500 + $2700

= $10200

5 0

3 years ago

Defining Terms

____ [38]

Answers: line DB

7 0

3 years ago

Adriana va a llenar bolsas de dulces para su fiesta de cumpleanos si tiene un paquete con 100 paletas y otro con 56 chocolates y

lions [1.4K]

Answer:

56 bags of a chocolate and a popsicle

Step-by-step explanation:

The first thing is to find the relationship between the number of popsicles and the number of chocolates, which would be the quotient of these:

100/56 = 1.78

Since it does not reach number two, you cannot add 2 popsicles and 2 chocolates, therefore the only way that they contain the same amount is that there is a chocolate and a popsicle, that is to say that a total of 56 bags would come out.

6 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