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Blababa [14]
2 years ago
7

Wiley is making monthly payments of $88.00 to pay off a loan that he took out to buy a fence, but he wants to pay off his loan f

aster. Which of these monthly payments will allow him to do so?
A. $96.00
B. $72.00
C. $64.00
D. $80.00
Mathematics
2 answers:
bekas [8.4K]2 years ago
6 0
Making larger monthly payments than required will pay off a loan faster.  Thus, the answer is A.
murzikaleks [220]2 years ago
6 0
$96.00 APEX !!!!!!!!!
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Consider the function f ( x ) = 2479 ⋅ 0.9948x First compare this with f ( x ) po ( 1 + r ) ^ 2 We get po = 2479 And 1 + r = 0.9948 = 1 – 0.0052 r = -0.0052 < 0 Therefore, f is an exponential decay function with a decay rate of 0.0052 x 100 = 0.52% 
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Emily is entering a bicycle race for charity. Her mother pledges $0.40 for every 0.25 mile she bikes. If Emily
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5 0
1 year ago
Find the dimensions of the box with volume 1728 cm3 that has minimal surface area. (Let x, y, and z be the dimensions of the box
Gala2k [10]

Answer:

The dimensions of the box are 12 cm , 12 cm , 12 cm

Step-by-step explanation:

Let x , y and z be the dimensions of box

Volume of box =xyz=1728

z=\frac{1728}{xy}

Surface area of box = 2xy+2yz+2xz=2xy+2y(\frac{1728}{xy})+2x(\frac{1728}{xy})

Let f(x,y)=2xy+2(\frac{1728}{x})+2(\frac{1728}{y})

To get minimal surface area

\frac{\partial f}{\partial x}=0 and \frac{\partial f}{\partial y}=0

\frac{\partial(2xy+2(\frac{1728}{x})+2(\frac{1728}{y}))}{\partial x}=0

2y-2(\frac{1728}{x^2})=0

y=\frac{1728}{x^2} ----1

\frac{\partial(2xy+2(\frac{1728}{x})+2(\frac{1728}{y}))}{\partial y}=0

2x-2(\frac{1728}{y^2})=0\\x=\frac{1728}{y^2}  \\y^2=\frac{1728}{x}

Using 1

(\frac{1728}{x^2} )^2=\frac{1728}{x}

x=0 and x^3=1728

Side can never be 0

So,x^3=1728

x=12

y=\frac{1728}{x^2} \\y=\frac{1728}{12^2}

y=12

z=\frac{1728}{xy}\\z=\frac{1728}{(12)(12)}

z=12

The dimensions of the box are 12 cm , 12 cm , 12 cm

5 0
2 years ago
-10√v-10=-60 Show All Steps.
docker41 [41]
-10rt(v-10)=-60
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5 0
1 year ago
Deal with these relations on the set of real numbers: R₁ = {(a, b) ∈ R² | a &gt; b}, the "greater than" relation, R₂ = {(a, b) ∈
uranmaximum [27]

Answer:

a) R1ºR1 = R1

b) R1ºR2 = R1

c) R1ºR3 = \{ (a,b) \in R^2 \}

d) R1ºR4 = \{ (a,b) \in R^2 \}

e) R1ºR5 = R1

f) R1ºR6 = \{ (a,b) \in R^2 \}

g) R2ºR3 = \{ (a,b) \in R^2 \}

h) R3ºR3 = R3

Step-by-step explanation:

R1ºR1

(<em>a,c</em>) is in R1ºR1 if there exists <em>b</em> such that (<em>a,b</em>) is in R1 and (<em>b,c</em>) is in R1. This means that a > b, and b > c. That can only happen if a > c. Therefore R1ºR1 = R1

R1ºR2  

This case is similar to the previous one. (<em>a,c</em>) is in R1ºR2 if there exists <em>b</em> such that (<em>a,b</em>) is in R2 and (<em>b,c</em>) is in R1. This means that a ≥ b, and b > c. That can only happen if a > c. Hence R1ºR2 = R1

R1ºR3

(a,c) is in R1ºR3 if there exists b such that a < b and b > c. Independently of which values we use for a and c, there always exist a value of b big enough so that b is bigger than both a and c, fulfilling the conditions. We conclude that any pair of real numbers are related.

R1ºR4

This is similar to the previous one. Independently of the values (a,c) we choose, there is always going to be a value b big enough such that a ≤ b and b > c. As a result any pair of real numbers are related.

R1ºR5

If a and c are related, then there exists b such that (a,b) is in R5 and (b,c) is in R1. Because of how R5 is defined, b must be equal to a. Therefore, (a,c) is in R1. This proves that R1ºR5 = R1

R1ºR6

The relation R6 is less restrictive than the relation R3, if we find 2 numbers, one smaller than the other, in particular we find 2 different numbers. If we had 2 numbers a and c, we can find a number b big enough such that a<b and b >c. In particular, b is different from a, so (a,b) is in R6 and (b,c) is in R1, which implies that (a,c) is in R1ºR6. Since we took 2 arbitrary numbers, then any pair of real numbers are related.

R2ºR3

This is similar to the case R1ºR3, only with the difference that we can take b to be equal to a as long as it is bigger than c. We conclude that any pair of real numbers are related.

R3ºR3

If a and c are real numbers such that there exist b fulfilling the relations a < b and b < c, then necessarily a < c. If a < c, then we can use any number in between as our b. Therefore R3ºR3 = R3

I hope you find this answer useful!

5 0
2 years ago
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