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2 years ago

Find the discriminant and the num x2 + 2x + 7 = 0 A. 28; two real roots

Mathematics

1 answer:

Helen [10]2 years ago

5 0

Answer:

Discriminant = -24, no real roots.

Step-by-step explanation:

The discriminant is what's inside the square root in the quadratic formula.

The quadratic formula is:

$\frac{-b+\sqrt{b^2-4ac} }{2a}$ and $\frac{-b-\sqrt{b^2-4ac} }{2a}$

a = 1

b = 2

c = 7

The question wants the discriminate, we can just focus on the square root part.

When we plug in the parts we got, it becomes:

$\sqrt{4-4(1)(7)}\\\sqrt{4-28} \\\sqrt{-24}$

Since the inside of the square root is negative, there are no real roots.

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Find the area of the figure.( sides meet at right angles.)picture below

Tema [17]

You can divide the figure as follows:

A vertical rectangle, 5 meters tall and 2 meters wide, on the top right.
A horizontal rectangle, 2 meters tall and 5 meters wide, on the bottom left
A 2x2 meters square on the bottom right, where the two rectangles meet.

The area of the rectangles is 5x2=10 meters squared, while the area of the square is 2x2=4 meters squared.

So, the total area is 10+4=14 meters squared.

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3 years ago

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a

Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q = -t/100 + c

$Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c} = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}$

when t = 0, Q = 200 L × 1 g/L = 200 g

$Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}$

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

$2 = 200e^{(-t/100)}\\\frac{2}{200} = e^{(-t/100)}$

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0

3 years ago

Is 2 greater or less than or equal to -5?

Marina CMI [18]

2 is greater then -5.

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