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MAVERICK [17]
3 years ago
13

The main difference between a secured loan and an unsecured loan is that the secured loan what?

Mathematics
1 answer:
Alex Ar [27]3 years ago
8 0
Includes a pledge of collateral
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a rectangle has a length that is nine feet less than four times its width. Its area is 90 ft. Algebraically determine the length
sammy [17]
Ok... after much confusion, I've come to the conclusion that you must have written the question wrong and it's actually the perimeter that is 90 feet so that's how I'm going to work it...

Width = w
Length = 4w - 9

P = 2w x 2L
90 = 2w + 2(4w - 9)
90 = 2w + 8w - 18
90 = 10w - 18
108 = 10w
108/10 = w
10 4/5 = w

Width is 10 4/5 ft

Length = 4w - 9
L = 4(10 4/5) - 9
L = 43 1/5 - 9
L = 34 1/5 ft

Length 34 1/5 ft  Width 10 4/5 ft or
Length 34.2 ft and width 10.8 ft

6 0
3 years ago
Which function is graphed below
Svetlanka [38]
I think the answer is a. f(x)= cos(x).
6 0
3 years ago
Read 2 more answers
4.) You have saved $2000 for college and are saving at the rate of $250 per month. Assume
Veseljchak [2.6K]

Answer: 5,000

Step-by-step explanation: 2,000 saved 250 per month so 250 times

7 0
3 years ago
3.The _________ of sets is the set of all the members of both sets without repeating any of the members in the sets.
KiRa [710]

Answer:

4.Ans: intersection part of set

6 0
3 years ago
A boat is pulled toward a dock by means of a rope wound on a drum that is located 6 ft above the bow of the boat. If the rope is
Monica [59]

Answer:

Step-by-step explanation:

Given that:

The height of the dock (h) = 6

Let represent d to be the distance between the boat and the dock

Let the length of the rope between the boat and the drum be denoted by (l)

Then, the rate of change for the length of the rope be:

dl/dt = -5 ft/s

Using Pythagoras rule to determine the relationship between these values, we have:

l^2 = h^2 +d^2

l^2 = 6^2 + d^2

l^2 = 36 + d^2

We relate to:  2l * \dfrac{dl}{dt} = 2d* \dfrac{dd}{dt}

From the question;

l = 34,

So to find \dfrac{dd}{dt}, we get;

d = \sqrt{l^2 - 36}

d = \sqrt{34^2 - 36}

d = \sqrt{1156- 36}

d = \sqrt{1120}

d = 33.46

So, we have:

\dfrac{dd}{dt}= \dfrac{l}{d} \times \dfrac{dl}{dt}

\dfrac{dd}{dt}= \dfrac{34}{33.46} \times - 5

\dfrac{dd}{dt}= 1.016 \times - 5

\dfrac{dd}{dt}=-5.08 \ ft/sec

4 0
3 years ago
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