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Delvig [45]
3 years ago
7

Write the coordinates of the vertices after a translation 6 units left and 2 units down.

Mathematics
1 answer:
jenyasd209 [6]3 years ago
4 0

Answer:

Answer below.

Step-by-step explanation:

K'(-7,0) L'(-5,3) M'(-7,4) N'(-9,3)

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What is 2a+b
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The answer is -8. Do you need an explanation?
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The perimeter of an equilateral triangle with side length s is 3s. Leslie is using a wooden equilateral triangle with 15-inch si
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2 years ago
A person invest $10,000 into a bank the bank pays 4.75% interest compounded semi annually. To the nearest 10th of a year, how lo
Mama L [17]

Answer:

T is 13.9 years to the nearest 10th of a year

Step-by-step explanation:

In this question, we are to calculate the number of years at which someone who invests a particular amount will have a particular amount based on compound interest.

To calculate the number of years, what we do is to use the compound interest formula.

Mathematically,

A = P(1+ r/n) ^nt

Where A is the final amount after compounding all interests which is $19,200 according to the question

P is the initial amount invested which is $10,000 according to the question

r is the rate which is 4.75% according to the question = 4.75/100 = 0.0475

n is the number of times per year in which interest is compounded. This is 2 as interest is compounded semi-annually

t= ?

we plug these values;

19200 = 10,000(1+0.0475/2)^2t

divide through by 10,000

1.92 = (1+0.02375)^2t

1.92 = (1.02375)^2t

We find the log of both sides

log 1.92 = log [(1.02375)^2t)

log 1.92 = 2tlog 1.02375

2t = log 1.92/log 1.02375

2t = 27.79

t = 27.79/2

t = 13.89 years

The question asks to give answer to the nearest tenth of a year and thus t = 13.9 years

7 0
3 years ago
At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is
Hitman42 [59]

Answer:

answer = 12.87 km/h

Step-by-step explanation:

Given

Ship A is sailing east at 25 km/h = \frac{dx}{dt}

ship B is sailing north at 20 km/h =\frac{dy}{dt}

here x and y are the  sailing at t = 4 : 00 pm for ship A and B respectively

so we get x = 4 ×25 =100 km/h

                 y = 4× 20 = 80 km/h

let z is the distance between the ships, we need to find \frac{dz}{dt} at t = 4 hr

At noon, ship A is 130 km west of ship B (12:00 pm)

so equation will be

z^2 = (130-x)^2 + y^2......................(i)\\put x = 100 and y = 80 \\\\we |  | get \\

z^2 = 30^2 + 80^2\\z =\sqrt{7300} km/h

derivative first equation w . r. to t we get

2z\frac{dz}{dt} =-2(130-x)\frac{dx}{dt}+2y\frac{dy}{dt}

\frac{dz}{dt} =\frac{1}{z}[(x -130)\frac{dx}{dt} +y\frac{dy}{dt}]

\frac{dz}{dt} = \frac{( -20\times25 + 80\times20)}{\sqrt{7300} }

     = \frac{1100}{85.44}\\  = 12.87km/h

8 0
3 years ago
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