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3 years ago

Sue invested $2400 at a 3% simple interest rate for 18 months. How much did she earn on her investment after the 18 months?

Mathematics

1 answer:

Anon25 [30]3 years ago

4 0

Answer:

sinple interest= principle*rate*time/100

SI= 2400*3*1.5/100

SI= 108

AMOUNT= SI + PRICIOLE

AMOUNT= 108+2400

AMOUNT= 2508

{ I converted 18 months to 1.5 years}

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Consider -q > 5. Use the addition and/or subtraction property of inequality to solve.

AleksandrR [38]

The correct answer would be -5 > q

In order to solve this using only addition and subtraction, what we need to do is change the side each term is on. This will allow us to get the variable to be a positive number.

-q > 5 ----> Subtract 5 from both sides

-5 - q > 0 -----> Add q to both sides

-5 > q

8 0

3 years ago

At what point does the curve have maximum curvature? Y = 4ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

MAXImum [283]

<u>Answer-</u>

At $x= \frac{1}{2304e^4-16e^2}$ the curve has maximum curvature.

<u>Solution-</u>

The formula for curvature =

$K(x)=\frac{{y}''}{(1+({y}')^2)^{\frac{3}{2}}}$

Here,

$y=4e^{x}$

Then,

${y}' = 4e^{x} \ and \ {y}''=4e^{x}$

Putting the values,

$K(x)=\frac{{4e^{x}}}{(1+(4e^{x})^2)^{\frac{3}{2}}} = \frac{{4e^{x}}}{(1+16e^{2x})^{\frac{3}{2}}}$

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

${k}'(x) = \frac{(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})}{(1+16e^{2x} )^{2}}$

Now, equating this to 0

$(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x}) =0$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}-(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}=(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{1}{2}}=48e^{2x}$

$\Rightarrow (1+16e^{2x})}=48^2e^{2x}=2304e^{2x}$

$\Rightarrow 2304e^{2x}-16e^{2x}-1=0$

Solving this eq,

we get $x= \frac{1}{2304e^4-16e^2}$

∴ At $x= \frac{1}{2304e^4-16e^2}$ the curvature is maximum.

6 0

2 years ago

HELP I CANT DO THIS-

Vera_Pavlovna [14]

1. C. Rectangle
2. F. 160
3. D. Triangular prism
4. H. 30 boxes
5. C. Equilateral Triangle

3 0

3 years ago

The vertices of a triangle are A (-2, 2), B (4,10), and C (0, -2). Find the equation of the line that contains

Tpy6a [65]

Answer:

y = 1/2 x + 3

Step-by-step explanation:

mid point of line BC = (2, 4)

slope of the line containing median (m) = 1/2

y intercept (b)= 3

since y = mx + b

y = 1/2x + 3

4 0

3 years ago

Please help and thank you

kondaur [170]

Answer:

Step-by-step explanation:

Let's see how well I can explain this. $\frac{\pi}{6}$ is the same as a 30 degree angle which is in quadrant 1. If you picture the unit circle, right in the center of it is the origin. If you draw a straight line from 30 degrees and through the center (the origin), you will automatically "connect" with the reference angle of 30 (this is true for ALL angles on the unit circle). This puts us in quadrant 3. In quadrant 3, x is negative and so is y. So the terminal point of the reference angle for 30 degrees has the same exact values, but both of them are negative (again, because both x and y are negative in quadrant 3). I can't see your choices but the one you want looks like this:

$(-\frac{\sqrt{3} }{2},-\frac{1}{2})$

3 0

2 years ago