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2 years ago

Please help!!

Mathematics

1 answer:

Igoryamba2 years ago

6 0

Answer:

Infinite solutions

Step-by-step explanation:

If you look at the equation, you will see that if you multiply everything by two (On the 1st equation), you get the 2nd equation. This means that the lines are the same, so infinite solutions.

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Alex Ar [27]

Answer:

ADB and ABC are similar

x = 8

Step-by-step explanation:

hypotenuse in ADB = $\sqrt{4^2 + 2^2 } = \sqrt{20}$

since two triangles are similar,

$\frac{2+x}{\sqrt{20} } = \frac{\sqrt{20}}{2}$

solving for x, x=8

4 0

3 years ago

Can someone please help me

djyliett [7]

Slope = (10 - 5)/(-9 + 15) = 5/6

y = mx + b where m = slope and b = y-intercept

b = y - mx

b = 5 - (5/6) (-15)

b = 5 +12.5

b = 17.5

Equation y = 5/6 x + 17.5

Y-intercept = 17.5

X-intercept when y = 0

5/6 x + 17.5 = 0

5/6 x = -17.5

x = (-17.5) (6/5)

x = -21

Answer

Y intercept (0, 17.5 )

X intercept (-21 , 0)

3 0

3 years ago

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lidiya [134]

Answer:

help* not gelp

Step-by-step explanation:

but i.d.k im so sry

6 0

2 years ago

Read 2 more answers

8/9+2/3 In fraction form

photoshop1234 [79]

8/9+6/9 equals 14/9 or 1 and 5/9

5 0

2 years ago

Read 2 more answers

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