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torisob [31]
3 years ago
7

Multiply 5x107 * 3x105

Mathematics
2 answers:
djverab [1.8K]3 years ago
8 0

Answer:

the first on 5×107=535 the second one 3×105=315

Bond [772]3 years ago
3 0

Answer:

5 x 107=535    3 x 105=315

Step-by-step explanation:

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Quinzel earns $1254 each month. His total deductions are 20% of his pay. How much is deducted from his pay each month ?
vova2212 [387]

20% = 0.20

1254 x 0.20 = 250.80

so he has $250.80 deducted every month


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3 years ago
at the cafeteria, muffins are sold in the following flavors: blueberry, orange, carrot, corn, bran, and choc chip. there is a re
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3 years ago
Mathematics algebra​
Alexxx [7]

Answer:

⇒  1\frac{1}{3}m-1\frac{1}{2}n-4\frac{1}{2}

Step-by-step explanation:

  • \frac{4m}{3}-\frac{3(n+3)}{2}
  • \frac{4m}{3}-\frac{3n+9}{2}
  • \frac{4m}{3}-\frac{3n}{2}-\frac{9}{2}
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5 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

2.1t=30

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
Review the diagram of a locket in the shape of an ellipse.
Helen [10]

Answer:

B

Step-by-step explanation:

If you graph y^2/361 + x^2/169=1 you will get the same graph depicted in the equation.

8 0
3 years ago
Read 2 more answers
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