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iren2701 [21]
2 years ago
14

Show that 7x + 1 is equivalence to 2(2x - 1) + 3(x + 1)​

Mathematics
2 answers:
marissa [1.9K]2 years ago
4 0
7x + 1 = 2(2x - 1) + 3(x + 1)

7x + 1 = 4x - 2 + 3x + 3

7x + 1 = 7x + 1

Please mark brainliest if right!
ryzh [129]2 years ago
3 0

Answer with Step-by-step explanation:

2 ( 2x -1 ) + 3 ( x + 1 )

First, let us <u>find the value of the above expression.</u>

2 ( 2x -1 ) + 3 ( x + 1 )

4x - 2 + 3x + 3

<em>Combine like terms</em>

4x + 3x - 2 + 3

<u>7x + 1</u>

Therefore, it is clear that the given expression is equal to 7x + 1.

∴ 7x + 1 = 2 ( 2x -1 ) + 3 ( x + 1 )

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a free tailed bat can fly 232 feet in 4 seconds. A hummingbird can fly 343 feet in 7 seconds. Which can fly fast and by how much
Katarina [22]

Step-by-step explanation:

First, we need to find the unit rate of each animal (how many feet each animal can fly in a second). We do this by dividing the number of feet flown by the number of time.

Bat: 232/4 = 58 ft per second
Hummingbird: 343/7 = 49 ft per second

Based off of this information, we can conclude that the bat can fly faster than the hummingbird.

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3 years ago
Having trouble with this, help please? <br><br> I will give brainliest.
yaroslaw [1]

Answer:

a)     7x-15=20

b)   \frac{20+15}{7}

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Step-by-step explanation:

4 0
2 years ago
A tank contains 10 liters of pure water. Saline solution with a variable concentration 5 grams of salt per liter is pumped into
algol [13]

Answer:

dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0

Step-by-step explanation:

The mass flow rate dQ(t)/dt = mass flowing in - mass flowing out

Since 5 g/L of salt is pumped in at a rate of 4 L/min, the mass flow in is thus 5 g/L × 4 L/min = 20 g/min.

Let Q(t) be the mass present at any time, t. The concentration at any time ,t is thus Q(t)/volume = Q(t)/10. Since water drains at a rate of 4 L/min, the mass flow out is thus, Q(t)/10 g/L × 4 L/min = 2Q(t)/5 g/min.

So, dQ(t)/dt = mass flowing in - mass flowing out

dQ(t)/dt = 20 g/min - 2Q(t)/5 g/min

Since the salt just begins to be pumped in, the initial mass of salt in the tank is zero. So Q(0) = 0

So, the initial value problem is thus

dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0

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2 years ago
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