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statuscvo [17]
4 years ago
11

What is half of the product of 8 1/7 and 7 4/5

Mathematics
1 answer:
UNO [17]4 years ago
6 0

First we must convert these mixed numbers to improper fractions:

8\frac{1}{7}=\frac{57}{7}

7\frac{4}{5}=\frac{39}{5}

So then we need the product of these two numbers. That means to multiply:

\frac{57}{7}* \frac{39}{5}= \frac{2,223}{35}

But then we need the value of half of this product. So then let's multiply by one-half:

\frac{2,223}{35} *\frac{1}{2} =\frac{2,223}{70}

Then we can convert this from an improper fraction to a mixed number:

\frac{2,223}{70} =31\frac{53}{70}

And so this number is half of the product of the initial two numbers.

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Pls find area of this shape
Basile [38]

Answer:

16+16

32

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
A woman walks due east on the deck of a ship at 4 mi/h. the ship is moving south at a speed of 24 mi/h. find the speed of the wo
grandymaker [24]

we are given

A woman walks due east on the deck of a ship at 4 mi/h.

the ship is moving south at a speed of 24 mi/h

so, firstly, we will draw diagram

so, we have

OW=4

OS=24

now, we can use Pythagoras theorem

SW^2=OS^2+OW^2

now, we can plug values

and we get

SW^2=24^2+4^2

SW=24.33

so,

the speed of the woman relative to the surface of the water is 24.33 mi/h .......Answer


6 0
3 years ago
What initial investment must be made to accumulate $60000 in 17 years if the money is invested in a mutual fund that pays 12% an
mars1129 [50]

$7881.18

Step-by-step explanation:

   Let the initial Investment be P_{0}. The Interest is compounded on a monthly basis at 12% annual interest rate. After 17 years, the Investment amounts to $60,000.

   As the annual interest rate is 12%, the monthly interest rate is 1%.

Since this is a compound interest problem, the total amount can be modeled as follows: P(t)=P_{0}(1+\frac{i}{100})^{t}

Here i is the interest rate, i.e 1, and t is the number of time periods, i.e 17\textrm{ years x }12\frac{\textrm{months}}{\textrm{year}}= 204\textrm{ months}

60,000=P_{0}\textrm{ x }(\frac{101}{100})^{204}

P_{0}=7881.18

∴ Initial Investment = $7881.18

4 0
3 years ago
Did I do this right?
Rasek [7]

Answer:

Yea u did......

Step-by-step explanation:

you did right because all the steps are right

5 0
3 years ago
Read 2 more answers
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
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