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anzhelika [568]
3 years ago
9

^{x} =16^{2/5}" alt="4^{3/4} * 2^{x} =16^{2/5}" align="absmiddle" class="latex-formula">
Mathematics
2 answers:
Sloan [31]3 years ago
5 0

Answer:

\sf x=\frac{1}{10}

Step-by-step explanation:

Rewrite expression with bases of 4.

\sf{4^{\frac{3}{4} }} \times \sf({4^\frac{1}{2} )^x =(4^2)^{\frac{2}{5} }

Apply law of exponents, when bases are same for exponents in multiplication, add the exponents. When a base with an exponent has a whole exponent, then multiply the two exponents.

\sf{4^{\frac{3}{4} }} \times \sf{4^{\frac{1}{2} x}=4^{\frac{4}{5} }

\sf{4^{\frac{3}{4} +\frac{1}{2} x}=4^{\frac{4}{5} }

Cancel same bases.

\sf \frac{3}{4} +\frac{1}{2} x=\frac{4}{5}

Subtract 3/4 from both sides.

\sf \frac{1}{2} x=\frac{1}{20}

Multiply both sides by 2.

\sf x=\frac{1}{10}

Salsk061 [2.6K]3 years ago
5 0

Step-by-step explanation:

2^{2*3/4} × 2^{x}=2^{4×2/5}

2^{3/2} × 2^{x}= 2^{8/5}

2^{3/2+x}=2^{8/5}

equate powers

{3+2x}/2= 2^2

5{3+2x}= 2{8}

15+10x=16

collect like terms

10x=16-15

10x=1

divide both sides by 10

x=1/10

x=0.1

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(20points + brainlesest) Tom has $10 that he wants to use to buy pens and pencils. however, he does not want to buy more pens an
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7 0
3 years ago
3. A rare species of aquatic insect was discovered in the Amazon rainforest. To protect the species, environmentalists declared
navik [9.2K]

The number of months until the insect population reaches 40 thousand is 14.29 months and the limiting factor on the insect population as time progresses is 250 thousands.

Given that population P(t) (in thousands) of insects in t months after being transplanted is P(t)=(50(1+0.05t))/(2+0.01t).

(a) Firstly, we will find the number of months until the insect population reaches 40 thousand by equating the given population expression with 40, we get

P(t)=40

(50(1+0.05t))/(2+0.01t)=40

Cross multiply both sides, we get

50(1+0.05t)=40(2+0.01t)

Apply the distributive property a(b+c)=ab+ac, we get

50+2.5t=80+0.4t

Subtract 0.4t and 50 from both sides, we get

50+2.5t-0.4t-50=80+0.4t-0.4t-50

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Divide both sides with 2.1, we get

t=14.29 months

(b) Now, we will find the limiting factor on the insect population as time progresses by taking limit on both sides with t→∞, we get

\begin{aligned}\lim_{t \rightarrow \infty}P(t)&=\lim_{t \rightarrow \infty}\frac{50(1+0.05t)}{2+0.01t}\\ &=\lim_{t \rightarrow \infty}\frac{50(\frac{1}{t}+0.05)}{\frac{2}{t}+0.01}\\ &=50\times \frac{0.05}{0.01}\\ &=250\end

(c) Further, we will sketch the graph of the function using the window 0≤t≤700 and 0≤p(t)≤700 as shown in the figure.

Hence, when the population P(t) (in thousands) of insects in t months after being transplanted by P(t)=(50(1+0.05t))/(2+0.01t) then the number of months until the insect population reaches 40 thousand 14.29 months and the limiting factor on the insect population is 250 thousand and the graph is shown in the figure.

Learn more about limiting factor from here brainly.com/question/18415071.

#SPJ1

8 0
2 years ago
Mitchell needs to take 2 medications. One
Archy [21]

Answer:

The 12th day.

Step-by-step explanation:

Every four days: 4th, 8th, 12th.  

Every three days: 3rd, 6th, 9th, 12th.

You just find lowest common multiple of four and three. Which just means the first multiple they share.

Hopes this makes sense!

6 0
3 years ago
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