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GuDViN [60]
2 years ago
5

Solve the following proportion for y.y/5 = 7/8Round your answer to the nearest tenth.

Mathematics
1 answer:
Vinvika [58]2 years ago
3 0

The given proportion is expressed as

y/5 = 7/8

If we crossmultiply, it becomes

y * 8 = 7 * 5

8y = 35

y = 35/8

y = 4.375

Rounding to the nearest tenth,

y = 4.4

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8x2 + 7x - 4 <br> what is the factor
astraxan [27]

Answer:

19

Step-by-step explanation:

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5 0
3 years ago
A town has 5000 people in year t = 0. Calculate how long it takes for the population P to double once, twice, and three times, a
UNO [17]

Answer:

Step-by-step explanation:

At the time t = 0, population of the town = 5000

Rate of population increase = 500 per year

Therefore, the equation that will represent the population will be

P_{t}=P_{0}+500t

Where P_{t} = Population after t years

P_{0}= Initial population

t = Time in years

a). For double once the population will be 500×2 = 10000

By plugging in the values in the equation,

10000 = 5000 + 500t

500t = 10000 - 5000

500t = 5000

t = \frac{5000}{500}

t = 10 years

For Double twice,

Population will be = 10000×2 = 20000

Now we plug in the values in the equation again

20000 = 5000 + 500t

500t = 20000 - 5000

500t = 15000

t = \frac{15000}{500}

t = 30 years

For double thrice,

Population of the town = 20000×2 = 40000

Now we plug in the values in the equation,

40000 = 5000 + 500t

500t = 40000 - 5000

500t = 35000

t = \frac{35000}{500}

t = 70 years

b). If the population growth is 5%.

Then the growth will be exponential represented by

T_{n}=T_{0}(1+\frac{r}{100})^{t}

T_{n} = Population after t years

T_{0} = Initial population

t = time in years

For double once,

Population after t years = 10000

10000=5000(1+\frac{5}{100})^{t}

(1.05)^{t}=\frac{10000}{5000}

(1.05)^{t}=2

Take log on both the sides

log(1.05)^{t}=log2

tlog(1.05) = log2

t = \frac{log2}{log1.05}

t = 14.20 years

For double twice,

Population after t years = 20000

20000=5000(1+\frac{5}{100})^{t}

(1.05)^{t}=\frac{20000}{5000}

(1.05)^{t}=4

Take log on both the sides

log(1.05)^{t}=log4

tlog(1.05) = log4

t = \frac{log4}{log1.05}

t = 28.413 years

For double thrice

Population after t years = 40000

40000=5000(1+\frac{5}{100})^{t}

(1.05)^{t}=\frac{40000}{5000}

(1.05)^{t}=8

Take log on both the sides

log(1.05)^{t}=log8

tlog(1.05) = log8

t = \frac{log8}{log1.05}

t = 42.620 years

4 0
3 years ago
The table below shows the surface area y in square inches, of a shrinking puddle in x hours
Kruka [31]
Part A
Correlation coefficient: -.99
This tells us that as time goes on (value of x increases) the area of the puddle goes down (value of y decreases)

Part B
y₂ - y₁
-------   = slope
x₂ - x₁
9 - 15
--------
5 - 8
-6/-3 = 2
So the slope equals -2, regardless of the fact that we got 2 as an answer there, we know that it is a negative slope 

Part C
The data represents causation because an increase in the value of x results in a decrease in the value of y, this shows an example of direct causation between x and y.
8 0
3 years ago
Read 2 more answers
A.) what was the scale factor of the dilation
Morgarella [4.7K]

Answer:

A. The scale factor is three.

B. Nine copies of ABCD could fit in A'B'C'D'.

Step-by-step explanation:

Hope this helps! Have a nice day! Let me know if you need an explanation!

3 0
3 years ago
It takes machine A x hours to manufacture a deck of cards that machine B can manufacture in y hours. If machine A operates alone
nlexa [21]

Answer:

The expression that shows how long the two machines will operate simultaneously is:

\frac{y(100x-z)}{x+y}

Step-by-step explanation:

We know that:

x: hours to manufacture a deck of cards for machine A

y: hours to manufacture a deck of cards for machine B

z: hours that machine A operates alone

The number of decks manufactured only by machine A is:

\frac{z}{x}

So, the remaining decks are given by:

100-\frac{z}{x}=\frac{100x-z}{x}

Then, the combined rate of machines A and B would be:

\frac{1}{x} +\frac{1}{y} =\frac{x+y}{xy}

The work-rate formula is:

Amount= Rate \times Time

Hence, the time that the two machines work simultaneously is:

Time=\frac{Amount}{Rate}

Time=\frac{Amount}{Rate} =\frac{\frac{100x-z}{x} }{\frac{x+y}{xy} } ={\frac{100x-z}{x} \times \frac{xy}{x+y}=\frac{y(100x-z)}{x+y}

3 0
4 years ago
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