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2 years ago

Pleease please please help me! asap <3 30 points! pls make sure its right UwU thx im timed :(

Mathematics

2 answers:

Andrej [43]2 years ago

8 0

Answer:

f(x) = -x -1

Step-by-step explanation:

f(x) = mx + b

find 2 points on the line:

A( -1 , 0)

B( 0 , -1)

Put them in the equation and fine m & b:

0=m(-1)+b

-1=m(0)+b

so b= -1

and m= -1

so f(x) = -x-1

maks197457 [2]2 years ago

8 0

Answer:

-x - 1

Step-by-step explanation:

line is passing through (-1, 0) (0, -1)

m = (-1 - 0) / (0 - (-1) )

= -1 / 1

= -1

line equation = y - y1 = m (x - x1)

y - 0 = -1 (x - (-1) )

y = -1 (x + 1)

y = -x - 1

y = f(x)

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How many inches longer is a hercules beetle than a ladybug

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Answer:

A Hercules Beetle is 6.97 inches larger than A ladybug

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A hercules beetle is 7 inches long while a ladybug is about 0.03 cm long when subtracting these two you will get a grand total of 6.97.

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2 years ago

Three balls are packaged in a cylindrical container as shown below. The balls just touch the top, bottom, and sides of the cylin

sp2606 [1]

Answer:

3,451 cm³

Step-by-step explanation:

The ball is a sphere.

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3 years ago

Maria's math test had 25 questions she got 84% correct how many did she get wrong

inysia [295]

$\frac{x}{25}=84% \\ \frac{x}{25}=.84 \\ 25( \frac{x}{25})=(.84)(25) \\ x=21$
Maria got 21 correct.

Number wrong = Total - Number right
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Hope this helps!

7 0

3 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

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago