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12345 [234]
3 years ago
11

Bill is a financial manager. He writes the equation A=2500(1.36)t to find out how much it will cost his company for a one-year l

oan of $2500 if the 36% APR is compounded only once. Which answer shows how the equation can be rewritten to find the interest rate that would cost Bill's company the same amount if it were compounded quarterly?
Mathematics
1 answer:
Viefleur [7K]3 years ago
4 0

Answer:

Step-by-step explanation:

First, to get the rate, we divide 36% by 100 to get point 36. Since the formula is A =P(1+Interest rate/ How many units its being compounded)^ Time x How many times its being compounded. It would make the equation 2500(1+.36/4)^Compounded once which is 1 x 4 = A.

Hope this helps.

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swat32
It’s d, 1 - t


t + 3 - 2 - 2t
we can rearrange it as
t - 2t + 3 - 2
-t + 1 is the same as 1 - t
5 0
3 years ago
Read 2 more answers
Consider the following sets of matrices: M2(R) is the set of all 2 x 2 real matrices; GL2(R) is the subset of M2(R) with non-zer
defon

Answer:

Step-by-step explanation:

REcall the following definition of induced operation.

Let * be a binary operation over a set S and H a subset of S. If for every a,b elements in H it happens that a*b is also in H, then the binary operation that is obtained by restricting * to H is called the induced operation.

So, according to this definition, we must show that given two matrices of the specific subset, the product is also in the subset.

For this problem, recall this property of the determinant. Given A,B matrices in Mn(R) then det(AB) = det(A)*det(B).

Case SL2(R):

Let A,B matrices in SL2(R). Then, det(A) and det(B) is different from zero. So

\text{det(AB)} = \text{det}(A)\text{det}(B)\neq 0.

So AB is also in SL2(R).

Case GL2(R):

Let A,B matrices in GL2(R). Then, det(A)= det(B)=1 is different from zero. So

\text{det(AB)} = \text{det}(A)\text{det}(B)=1\cdot 1 = 1.

So AB is also in GL2(R).

With these, we have proved that the matrix multiplication over SL2(R) and GL2(R) is an induced operation from the matrix multiplication over M2(R).

7 0
3 years ago
Ecuación 2:<br> 10x + 20y = 8000<br> ayudaaa
Elenna [48]

Answer:

y = 300

x = 200

Step-by-step explanation:

Esta pregunta parece incompleta, parece que acá tenemos un sistema de ecuaciones:

x + y = 500

10*x + 20*y = 8000

Estas ecuaciones tienen que resolverse en conjunto, y de acá sacaremos un par de puntos (x, y) que son solución para ambas ecuaciones.

El primer paso para resolver esto es aislar una variable en una de las ecuaciones, en este caso podemos aislar x en la primera ecuación y así obtener:

x = 500 - y.

Ahora podemos reemplazar eso en la segunda ecuación y obtener:

10*(500 - y) + 20*y = 8000.

Ahora resolvemos esto para y.

5000 - 10*y + 20*y = 8000

10*y = 8000 - 5000 = 3000

y = 3000/10 = 300.

Y sabíamos que:

x = 500 - y = 500 - 300 = 200

Entonces la solución es:

y = 300

x = 200

4 0
3 years ago
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IrinaVladis [17]
8x - 6 - 12x^2 + 9x

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=> (2-3x)(3x-3)

=> 3(x-1)(2-3x)
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Help, I am unsure about what to do exactly
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You plug in number for number getting 194.8 which is equal to 2931
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