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3 years ago

$1,030 at 4% compounded semiannually for 2 years

Mathematics

1 answer:

tatyana61 [14]3 years ago

6 0

Answer:

A=$1, 114.91 is the answer. Hope this helps!

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En una tienda se anuncia la rebaja de 20% en todos sus artículos . Rafael elije una camisa y al llegar a la caja le informan que

dimulka [17.4K]

Responder:

$ 750

Explicación paso a paso:

Dado que :

Descuento en todos los artículos = 20%

Descuento adicional dado = 15%

Monto final pagado por la camisa = $ 510

Precio de la camiseta sin descuento:

Sea el precio inicial de la camisa = x

Primer descuento obtenido = 20%

Precio pagado después de un descuento adicional del 15%

Primer descuento:

(1 - 0,20) * x = 0,8x

Segundo descuento:

(1 - 0,15) * 0,8x = 510

0,85 * 0,8x = 510

0,68x = 510

x = 510 / 0,68

x = $ 750

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3 years ago

How long does it take to reach 2000$ with an investment of 500$ with a annual interest rate of 6% and doubles every 12 years?

In-s [12.5K]

First you add 2000 and 500 and then when you get the answer you add 6 and 12 then you get the answer

3 0

3 years ago

How do you solve this URGENT (50pts)

Aliun [14]

Explanation:

x + 3y = 2

Converting it to slope intercept form:

y = mx + b [where m is slope, b is y-intercept]

Make y the subject:

$\sf \rightarrow 3y = 2 - x$

$\sf \rightarrow 3y = -x + 2$

$\sf \rightarrow y = -\dfrac{x}{3} + \dfrac{2}{3}$

$\sf \rightarrow y = -\dfrac{1}{3} x+ \dfrac{2}{3}$

which reveals slope:

$\star \ slope: \sf - \dfrac{1}{3}$

7 0

2 years ago

Read 2 more answers

Which of the following best describes the term induction?

sergey [27]

Reducing the solution to a problem in lowest terms

7 0

3 years ago

Before a bag of flour can be sold, it must be 40 kilograms but can be within 0.5

Serjik [45]

Answer:

The minimum acceptable weight of the flour bag is 39.5 kg,

and its maximum acceptable weight is 40.5 kg

Step-by-step explanation:

Notice that the selling weight of the bag to be sold (x) needs to differ from 40 kg at most 0.5 kg. Therefore, we can write the following inequality:

$|x-40|\leq 0.5$

The inequality can be solve once we remove the absolute value symbol, and solve for the unknown "x". Recall that in order to solve it, we need to consider the two possible cases:

a) that the expression within the absolute value symbol is larger than or equal to zero, and b) that the expression is less than zero.

a) If $x-40\geq 0$ then its absolute value is equal to itself (x-40), and when we remove the absolute value symbols, we get:

$x-40\leq 0.5\\x\leq 0.5+40\\x\leq 40.5\,\,kg$

which means that the weigh "x" of the flour bag should be smaller or equal than 40.5 kg

b) If $x-40$ , then the absolute value of this is its opposite : -x+40, and the inequality becomes:

$-x+40\leq 0.5\\40\leq 0.5+x\\40-0.5\leq x\\39.5\leq x$

which means that the weight of the flour bag must be larger than or equal to 39.5 kg

So, the minimum acceptable weight of the flour bag is 39.5 kg, and the maximum acceptable weight is 40.5 kg

6 0

3 years ago