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inysia [295]
2 years ago
12

Rick discounts a 100-day note for $25,000 at 13%. The effective rate of interest to the nearest hundredth percent is

Mathematics
1 answer:
zhuklara [117]2 years ago
3 0
First we have to calculate the amount of interest. This can be done by using the interest formula, I = P x R x T.

= 25,000 x 0.13 x (100/365)
= 902.78

This will be deducted from the principal amount, which is 25,000. 

= 24,097

Last, we plug our values in to the effective rate formula for interest: 

\frac{Interest}{Proceeds x Time of Note}

902.78/6693.6722

= 13.487066

After rounding, we come to our final answer, which is D, 13.49%. 
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Select all the properties that are used below to solve 7y − 15 = −29. 7y – 15 + 15 = –29 + 15 7y = –14 7y7 = –147 y = –2 A. Iden
lukranit [14]

B.addition property of multiplication

D.inverse property of multiplication

E.commutative property of addition

3 0
2 years ago
Find the real numbers x and y if -3+ix^2y and x^2+y+4i are conjugate of each other. Pls solve with the steps
Firdavs [7]
ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work

EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same

For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y

For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4

Therefore, for the two expressions to be conjugates, we must satisfy the two conditions. 

Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the 

   x²y = -4 ... (I)

Condition 2: Real parts are the same

   x² + y = -3 ... (II)

We have a system of equations since both conditions must be satisfied

   x²y = -4 ... (I)
   x² + y = -3 ... (II)

We can rearrange equation (II) so that we have

   y = -3 - x² ... (II)

Substituting into equation (I)

   x²y = -4 ... (I)
   x²(-3 - x²) = -4
   -3x² - x⁴ = -4
   x⁴ + 3x² - 4 = 0
   (x² + 4)(x² - 1) = 0
   (x² + 4)(x-1)(x+1) = 0

Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.

Solve for y:

   y = -3 - x² ... (II)
   y = -3 - (±1)²
   y = -3 - 1
   y = -4

So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:

   -3 + ix²y 
   = -3 + i(±1)²(-4)
   = -3 - 4i

   x² + y + 4i
   = (±1)² - 4 + 4i
   = 1 - 4 + 4i
   = -3 + 4i

They result in conjugates
4 0
2 years ago
Read 2 more answers
How do I factorise 35x+55
densk [106]
Find what is common between 35 and 55....and that would be 5....so factor 5 out

35x + 55 =
5(7x + 11) <==


5 0
2 years ago
Read 2 more answers
On a coordinate plane, an absolute value graph starts at (0, 0) and goes up and to the left through (negative 4, 2). The functio
natka813 [3]

Answer:

The function f(x)= –StartRoot negative x EndRoot is shown on the graph.

On a coordinate plane, an absolute value graph starts at (0, 0) and goes down and to the left through (negative 4, negative 2).

Step-by-step explanation:

The range of the graph is all real numbers greater than or equal to 0.

9 0
2 years ago
Read 2 more answers
BRAINLIESTTT ASAP!! PLEASE HELP ME :)
Elanso [62]

Answer:

x=-2 (the actual solution)

x=-5 (the extraneous solution)

Step-by-step explanation:

First step is to isolate the radical part (the part with the square root).

So we will add 4 on both sides first:

\sqrt{x+6}=x+4

To get rid of square we are going to square both sides:

x+6=(x+4)^2

We are going to expand right hand side (you can use "F.O.I.L":

x+6=x^2+4x+4x+16

Subtract x on both sides and subtract 6 on both sides:

0=x^2+4x+4x-x+16-6

Combine like terms:

0=x^2+7x+10

We will now factor:

0=(x+2)(x+5) (Since 2(5)=10 and 2+5=7)

Set both factors equal to 0:

x+2=0 or x+5=0

Solve first equation by subtracting 2 on both sides to obtain x=-2.

Solve second equation by subtracting 5 on both sides to obtain x=-5.

Now let's check to see if these are both actually indeed solutions to the equation we began with:

\sqrt{x+6}-4=x for x=-2:

\sqrt{-2+6}-4=-2

\sqrt{4}-4=-2

2-4=-2

-2=-2 is a true equation so x=-2 is indeed a solution.

\sqrt{x+6}-4=x for x=-5:

\sqrt{-5+6}-4=-5

\sqrt{1}-4=-5

1-4=-5

-3=-5 is false so x=-5 is what we call the extraneous solution.

5 0
2 years ago
Read 2 more answers
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