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4 years ago

A savings account earns 5% simple interest per year. The principal is $1200. What is the balance after 4 years?

2 answers:

7 0

The formula of the Simple Interest is:
I=PRT
P for Principle Amount ($1200)
R for Rate (5%= $\frac{5}{100}$ = 0.05)
T for Time in years (4 years)
I = 1200 × 0.05 × 4
= $240

Add the interest to the principle amount to check the balance
$1200 + $240 = $1440

irina1246 [14]4 years ago

7 0

$240 i rember doing interest right around january

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Step-by-step explanation:

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Suppose a parabola has vertex (6,5) and also passes through the point (7,7). Write the equation of the parabola in vertex form.

jek_recluse [69]

Answer:

Choice B: $y = 2\, (x - 6)^{2} + 5$ .

Step-by-step explanation:

For a parabola with vertex $(h,\, k)$ , the vertex form equation of that parabola in would be:

$\text{$y = a\, (x - h)^{2} + k$ for some constant $a$}$ .

In this question, the vertex is $(6,\, 5)$ , such that $h = 6$ and $k = 5$ . There would exist a constant $a$ such that the equation of this parabola would be:

$y = a\, (x - 6)^{2} + 5$ .

The next step is to find the value of the constant $a$ .

Given that this parabola includes the point $(7,\, 7)$ , $x = 7$ and $y = 7$ would need to satisfy the equation of this parabola, $y = a\, (x - 6)^{2} + 5$ .

Substitute these two values into the equation for this parabola:

$7 = (7 - 6)^{2}\, a + 5$ .

Solve this equation for $a$ :

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Hence, the equation of this parabola would be:

$y = 2\, (x - 6)^{2} + 5$ .

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Answer:

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Step-by-step explanation:

You can refer to the picture.

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So, the angle lies in quadrant 2 (Sin quadrant)

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3 years ago

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Murljashka [212]

Answer:

$\boxed{\pink{\tt I = \dfrac{1}{3}sin(3x) - 3cos(x) + C}}$

Step-by-step explanation:

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$\implies\displaystyle\sf I = \int cos\ u \dfrac{1}{3}du + \int 3sin \ x \ dx \\\\\implies\displaystyle \sf I = \int \dfrac{cos\ u}{3} du + \int 3sin\ x \ dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3}\int \dfrac{cos(u)}{3} + \int 3sin(x) dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3} sin(u) + C +\int 3sin(x) dx \\\\\implies\displaystyle \sf I = \dfrac{1}{3}sin(u) + C + 3\int sin(x) \ dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3}sin(u) + C + 3(-cos(x)+C) \\\\\implies \underset{\blue{\sf Required\ Answer }}{\underbrace{\boxed{\boxed{\displaystyle\red{\sf I = \dfrac{1}{3}sin(3x) - 3cos(x) + C }}}}}$

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