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OverLord2011 [107]
3 years ago
5

If Millie invests $35,000 in a savings account that earns 3.5% interest, how much interest will she earn after 3 years?

Mathematics
1 answer:
deff fn [24]3 years ago
3 0

Answer:

$3,675

Step-by-step explanation:

Simple Interest = principal x rate x time

so, simple interest = $35,000 x 3.5/100 x 3

Therefore, the answer is $3,675

Hooe this helped :)

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6 0
3 years ago
PLEASE help me with this question! This is really urgent! No nonsense answers please.
ipn [44]

Answer:

The second table of values.

Step-by-step explanation:

Let's put the x-values in the second table of values in correct number order:

x: -3, -2, -1, 0, 1

Now, let's write out the y-values in correct number order:

y: 1/4, 1, 4, 16, 64

Finally, let's rewrite the second table of values with the x-values in order and the corresponding y-values underneathe:

x: -3,  -2,  -1  0  1

y: 64,  16,  4, 1,  1/4

As it can be seen, as the x-values get bigger in value, the y-values get smaller exponentially, which is the definition of exponential decay.

6 0
3 years ago
Find all polar coordinates of point p where p = (6, -pi/5)
Lelu [443]
Given a point in coordinates form (r,\theta), one can compute the cartesian form like this:
(r\cos\theta,r\sin\theta )
We have:
r=6,\theta=-\dfrac{\pi}{5}
We get the cartesian form then:
\left(6\cos-\dfrac{\pi}{5},6\sin-\dfrac{\pi}{5}\right)
4 0
3 years ago
Help please...........
Snezhnost [94]

Answer:

66.42° for Y

23.58° for Z

90° for X

4 0
3 years ago
Find the derivative of the function at P 0 in the direction of A. ​f(x,y,z) = 3 e^x cos(yz)​, P0 (0, 0, 0), A = - i + 2 j + 3k
Alik [6]

The derivative of f(x,y,z) at a point p_0=(x_0,y_0,z_0) in the direction of a vector \vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k is

\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}

We have

f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k

and

\vec a=-\vec\imath+2\,\vec\jmath+3\,\vec k\implies\|\vec a\|=\sqrt{(-1)^2+2^2+3^2}=\sqrt{14}

Then the derivative at p_0 in the direction of \vec a is

3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}

3 0
4 years ago
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