All categories

3 years ago

Suppose it investment of $82,000 dollars in value every seven years. how much is the investment worth after 28 years

1 answer:

8 0

The answer is $328,000. Since the investment is $82,000 every seven years, and there is 28 years. If you divide 28 by 7 you get 4. Then you multiply 82,000 by 4 to get your answer : $328,000.

You might be interested in

Two interior angles of a scalene triangle measure 85 and 50 degrees. The third angle measures:

zmey [24]

The interior of a triangle = 180°
So 180° - 85° -50° = 45°

c. 45 degrees

3 0

2 years ago

Can someone help me with this answer plsss

solniwko [45]

Answer: 44 ft

Step-by-step explanation:

if you add up the right square it adds up to 16 4+4+4+4 all sides are equal since its a square.

The left rectangle is

8+8=16

6+6=12

16+12=28

Now you add the square on the right to the square on the left which is

28+16= 44 ft

4 0

2 years ago

Plz, help!!!!! 15 pts

Korolek [52]

3/12 + 10/12 = 13/12 because the points lie in-between the 1/6 and 2/6 marks and the 6/6 and 7/6 marks.

6 0

3 years ago

Read 2 more answers

-3x+6=21 -3x+6<21 which statement best describes the process used to slove the equations

sukhopar [10]

Answer:

-3x-3x=21-6

9x=15

x=15/9

5 0

2 years ago

Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

finlep [7]

Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

b) $a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

7 0

3 years ago