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3 years ago

13

Jason bought a new fish aquarium (shown below) and filled it with water half way. What is the volume of water that Jason put in

his aquarium?

1 answer:

Doss [256]3 years ago

6 0

Answer:

Step-by-step explanation:

The volume is Halfway

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PLEASE PLEASE PLEASE HELP ME

qaws [65]

Answer:

$\large\boxed{\dfrac{1}{45}}$

Step-by-step explanation:

1.

2 yellow marables of 10 all marables

$P(A_1)=\dfrac{2}{10}=\dfrac{1}{5}$

2.

1 yellow marable of 9 all marables

$P(A_2)=\dfrac{1}{9}$

$P(A)=P(A_1)\cdot P(A_2)\\\\P(A)=\dfrac{1}{5}\cdot\dfrac{1}{9}=\dfrac{1}{45}$

5 0

2 years ago

Solve the separable differential equation dtdx=x2+164 and find the particular solution satisfying the initial condition x(0)=9

maria [59]

$\dfrac{dt}{dx} = x^2 + \frac{1}{64} \Rightarrow\ dt = \left(x^2 + \frac{1}{64}\right)dx \Rightarrow \\ \\ \displaystyle \int 1 dt = \int \left(x^2 + \frac{1}{64}\right)dx \Rightarrow \\ t = \dfrac{x^3}{3} + \frac{x}{64} + C$

C = 9 because all the x terms go away.

$t = \dfrac{x^3}{3} + \dfrac{x}{64} + 9$

3 0

3 years ago

You open a drink stand where you sell coffee and hot chocolate. You sell coffee for $1.80 and hot chocolate for $3.15. At the en

Usimov [2.4K]

Answer:

c = 69 and h = 51

Step-by-step explanation:

You will need two different equations here

use variable c as coffee and variable h as hot chocolate

c + h = 120

1.80c + 3.15h = 284.40

By moving variables, you can alter the first equation

h = 120 - c

Now insert this into the second equation

1.80c + 3.15(120 - c) = 284.40

Multiply

1.80c + 378 - 3.15c = 284.40

Subtract the c

-1.35c + 378 = 284.40

Subtract 378 from both sides

-1.35c = -93.6

Divide both sides by -1.35

c = 69

Then plug into the first equation

69 + h = 120

Subtract 69

h = 51

7 0

2 years ago

24. q + r t s, for q = 46, r = 19. s = 54

Kaylis [27]

$(24 \times 46 )+ (19 \times t \times 54) =$

$1104 + 1026t$

_________________________________

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

8 0

3 years ago

Show that the transformation T defined by<br>T(x1, x2) = (2x1 - 3x2, X1+4, 5x2) is not linear.<br>

erastovalidia [21]

Let <em>u</em> = (1, 0) and <em>v</em> = (0, 1). Then

<em>T</em> (<em>u</em>) = (2*1 - 3*0, 1 + 4, 5*0) = (2, 5, 0)

<em>T</em> (<em>v</em>) = (2*0 - 3*1, 0 + 4, 5*1) = (-3, 4, 5)

=> <em>T</em> (<em>u</em>) + <em>T</em> (<em>v</em>) = (-1, 9, 5)

but

<em>T</em> (<em>u</em> + <em>v</em>) = <em>T</em> (1, 1) = (2*1 - 3*1, 1 + 4, 5*1) = (-1, 5, 5)

=> <em>T</em> (<em>u</em> + <em>v</em>) ≠ <em>T</em> (<em>u</em>) + <em>T</em> (<em>v</em>)

which means <em>T</em> does not preserve addition, so it is not linear.

4 0

2 years ago