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yKpoI14uk [10]
3 years ago
15

Felicia keeps track of the changes in value of her baseball card collection.for what three situations might the changes be repre

sented by a positive number,a negative number, and 0?
Mathematics
1 answer:
lord [1]3 years ago
6 0
If she gains a card that would be a positive 1, if she sells one that would be negative 1, and if she keeps one than that would be 0.
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Factorise<br> 24p2 + pq - 23q²​
Novosadov [1.4K]

<em>24p² + pq - 23q² = </em>

<em>= 24p² + 24pq - 23pq - 23q²</em>

<em>= 24p(p + q) - 23q(p + q)</em>

<em>= (p + q)(24p - 23q)</em>

<em />

<em />

3 0
3 years ago
I really need sum help plz
timama [110]

Answer:

32 Strawberries

Step-by-step explanation:

The amount of strawberries you need for 48 cups of juice is 32

3 0
2 years ago
Solve by substitution 2.5x-3y=-3 3.25x-y=-14
zhuklara [117]
Hello Complimentmady1, the first question, answer: <span>x=<span><span>1.2y</span>−1.2</span></span><span> second question, </span>answer:<span /><span> x=<span><span>0.307692y</span>−4.307692.</span></span>


4 0
3 years ago
Find the mode.<br>58, 51, 54, 49, 59, 53, 55, 58, 41, 44​
boyakko [2]
When you are finding the mode your looking for the number that shows up the most. In this the mode is 58.
8 0
3 years ago
Read 2 more answers
Your swimming pool containing 60,000 gal of water has been contaminated by 6 kg of a nontoxic dye that leaves a swimmer's skin a
Paul [167]

[a] Dye is removed from the pool at a rate of

(250 gal/min) * (<em>q</em>/60,000 g/gal) = -<em>q</em>/240 g/gal

where <em>q</em> denotes the amount of dye in the pool at time <em>t</em>. Clean water is pumped back into the pool, so no dye is being re-added.

So the net rate of change of the amount of dye in the pool is given by the differential equation,

\dfrac{\mathrm dq}{\mathrm dt}=-\dfrac{q(t)}{240}

with the intial value, <em>q</em>(0) = 6000 g (or 6 kg).

[b] The ODE above is separable as

\dfrac{\mathrm dq}q=-\dfrac{\mathrm dt}{240}

Integrate both sides to get

\ln|q|=-\dfrac t{240}+C

e^{\ln|q|}=e^{-t/240+C}

\implies q(t)=e^{-t/240+C}=e^{-t/240}e^C=Ce^{-t/240}

Now plug in the initial condition:

6000=Ce^0\implies C=6000

so the particular solution to the IVP is

q(t)=6000e^{-t/240}

[c] The acceptable concentration of the dy is 0.03 g/gal, which in a pool containing 60,000 gal of water corresponds to

(0.03 g/gal) * (60,000 gal) = 1800 g = 1.8 kg

of dye. Find the time <em>t</em> when this occurs:

1800=6000e^{-t/240}\implies0.3=e^{-t/240}

\implies\ln0.3=-\dfrac t{240}

\implies t=-240\ln0.3\approx288.953

so the amount of dye in the pool is within the acceptable tolerance after about 289 min have passes, or about 4.82 hrs. So no, the filtration system is not up to the task.

8 0
3 years ago
Read 2 more answers
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