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murzikaleks [220]

3 years ago

12

I need this answer (IMPORTANT) i'm not so good at math

2 answers:

MakcuM [25]3 years ago

7 0

Answer:

G- 3 cubic

Step-by-step explanation:

nordsb [41]3 years ago

6 0

I would say “G. 3 Cubic Centimeters”

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Determine if the two triangles are congruent. If they are, state how you know.

enot [183]

Answer:

C

Step-by-step explanation:

the hypotenuse and the leg of the first right angle are congruent to the other triangle's.

6 0

3 years ago

Please help the teacher doesn’t teach and i just need to pass this class

Aloiza [94]

ANSWER IS 73 DEGREE

BECAUSE IF WE WILL ADD 42 AND 65 WE WILL GET 107

THEN WE HAVE TO SUBTRACT 180 FROM 107

WE WILL GET 73

HOPE IT WILL HELP YOU

3 0

3 years ago

A large tank is filled to capacity with 300 gallons of pure water. Brine containing 5 pounds of salt per gallon is pumped into t

Gwar [14]

Answer:

(a) $A = 1500(1 - e^{-0.01t})$

(b) $t = 0$

Step-by-step explanation:

Given

$V= 300gal$ --- Volume of tank

$B = 5lb/gal$ --- Brine solution

$R_1 = 3gal/min$ --- Rate in, in gallon/min

$R_2 = 6gal/min$ --- Rate out, in gallon/min

Required

Determine A(t)

First, calculate the rate in (R in) and (R out) in lb/min

$R_{in} = B * R_1$

$R_{in} = 5lb/gal * 3gal/min$

$R_{in} = 15lb/min$

R(out) is calculated by multiply the rate at which brine leaves the tank (lb/gal) * R2

So, we have:

$R_{out} = \frac{A(t)}{300} lb/gal * 3gal/min$

$R_{out} = \frac{3*A(t)}{300} lb/min$

$R_{out} = \frac{A(t)}{100} lb/min$

The change in the amount of brin e in the tank at time t is given as:

$A'(t) = R_{in} - R_{out}$

$A'(t) = 15 - \frac{A(t)}{100}$

Rewrite:

$\frac{dA}{dt} = 15 - \frac{A}{100}$

Multiply through by dt

$dA = [15 - \frac{A}{100}]* dt$

Make dt stand alone

$\frac{dA}{15 - \frac{A}{100}} = dt$

$ln|15 - \frac{A}{100}| = -0.01t + ln\ c$

Express as exponents

$15 - \frac{A}{100} = ce^{-\frac{t}{100}}$

Multiply through by 100

$1500 - A = 100ce^{-\frac{t}{100}}$

$A = 1500 - 100ce^{-\frac{t}{100}}$

$A = 1500 - 100ce^{-0.01t}$

Apply initial conditions

$A(0) = 0$

Substitute 0 for t in: $A = 1500 - 100ce^{-0.01t}$

$0 = 1500 - 100ce^{-0.01*0}$

$0 = 1500 - 100ce^{0}$

$0 = 1500 - 100c$

$100c = 1500$

$c = 15$

Substitute 15 for c in $A = 1500 - 100ce^{-0.01t}$

$A = 1500 - 100*15e^{-0.01t}$

$A = 1500 - 1500e^{-0.01t}$

Factorize:

$A = 1500(1 - e^{-0.01t})$

How long to empty the tank

Set A to 0

$1500(1 - e^{-0.01t}) = 0$

Divide through by 1500

$1 - e^{-0.01t} = 0$

Collect Like Terms

$e^{-0.01t} = 1$

Take ln of both sides

$ln\ (e^{-0.01t}) = ln\ 1$

$-0.0t = 0$

$t = 0$

8 0

3 years ago

Help solve -4 + n/2 = -12

Nat2105 [25]

Answer:

n=-16

Step-by-step explanation:

Add 4 on each side:

n/2 = -12+4

n/2 = -8

Multiply 2 on each side:

n=-16

6 0

3 years ago

Describe which of the two points lies on the graph of the line<br> Y-x=4<br> A. 9,5<br> B. 5,9

klasskru [66]

B. 5, 9

x = 5
y = 9

y - x = 4

(9) - (5) = 4

True

hope this helps

4 0

3 years ago