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1 year ago

600 can be written as 2^a x b x c^d. where a, b, c and d are all prime numbers. find the values of a, b, c and d

Mathematics

1 answer:

Sergeu [11.5K]1 year ago

8 0

Answer:

easy one

Step-by-step explanation:

first factorise 600 as 2*2*2*3*5*5

then write as 2^3 * 3*5^2= 2^a*b*c^d

equating corresponding elements as a=3

b=1and c= 2 is the answer

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A new car is purchased for 15300 dollars. The value of the car depreciates at 14.25% per year. What will the value of the car be

Andrei [34K]

Answer:

$\$6,082.70$

Step-by-step explanation:

we know that

The formula to calculate the depreciated value is equal to

$V=P(1-r)^{x}$

where

V is the depreciated value

P is the original value

r is the rate of depreciation in decimal

x is Number of Time Periods

in this problem we have

$P=\$15,300\\r=14.25\%=0.1425\\x=6\ years$

substitute in the formula

$V=15,300(1-0.1425)^{6}$

$V=15,300(0.8575)^{6}$

$V=\$6,082.70$

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2 years ago

A person is watching a boat from the top of a lighthouse. The boat is approaching the lighthouse directly. When first noticed, t

maksim [4K]

Answer:

Boat traveled 553.24 feet towards the lighthouse.

Step-by-step explanation:

In the figure attached AB is the light house of height 200 feet.

Angle of depression of the boat from the top of a lighthouse = angle of elevation of the lighthouse from the boat = 14°52'

so 1' = $\frac{1}{60}$ degree

so angle of elevation at point C = 14 + $\frac{52}{60}$

So angle of elevation from C = (14 + 0.87) = 14.87°

Similarly, when boat arrives at point D angle of elevation = 45°10' = 45 + $\frac{10}{60}$ = 45.17°

Now we have to calculate the distance CD, traveled by the boat.

In ΔABC

tan14.87 = $\frac{200}{BC}$

0.2655 = $\frac{200}{BC}$

BC = $\frac{200}{0.26552}$

BC = 753.239 feet

Similarly in ΔABD

tan45.17 = $\frac{200}{BD}$

1 = $\frac{200}{BD}$

BD = 200 feet

So distance CD = BC - BD

CD = 753.239 - 200

= 553.24 feet

Therefore, Boat traveled 553.24 feet towards the lighthouse.

7 0

2 years ago

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A length is measured as 21cm correct to 2 significant figures. what is the upper bound for the length?

Leno4ka [110]

the upper bound for the length is $21.5 cm$ .

<u>Step-by-step explanation:</u>

Lower and Upper Bounds

The lower bound is the smallest value that will round up to the approximate value.
The upper bound is the smallest value that will round up to the next approximate value.

Ex:- a mass of 70 kg, rounded to the nearest 10 kg, The upper bound is 75 kg, because 75 kg is the smallest mass that would round up to 80kg.

Here , A length is measured as 21cm correct to 2 significant figures. We need to find what is the upper bound for the length . let's find out:

As discussed above , upper bound for any number will be the smallest value in decimals which will round up to next integer value . So , for 21 :

⇒ $21.5 cm$

21.5 cm on rounding off will give 22 cm . So , the upper bound for the length is $21.5 cm$ .

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2 years ago

You work for a landscaper that has a customer needing to seed an area of land 80 feet by 40 feet in size. The garden center has

Andreas93 [3]

You need to convert your 3200 sq ft into yards which is 355.556 sq yards then you divide 355.556 by 25 to get 14.22224 bags of seed needed

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2 years ago

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Marina CMI [18]

Answer: y=12 x=28

Explanation:
x=Amount of Small Lanterns y=Amount of Large Lanterns
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x+y=40
x=40-y
25(40-y)+40y=1180
1000-25y+40y=1180
1000+15y=1180
15y=180
y=12
x+12=40
x=28

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2 years ago

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