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2 years ago

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You deposit $300 in a savings account the account earns 2% simple interest per year. What is the interest earned after 10 years?

What is the balance after 10 years?

1 answer:

natita [175]2 years ago

6 0

Answer:kkjkhj

Step-by-step explanation:

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Please help me solve this problem... I will mark you brainliest

DaniilM [7]

Answer:

As the wheel makes this 270 degree counterclockwise rotation about the origin, the y-coordinate of the first car decreases from 80 to 0 and then further from 0 to -80, and finally increases to 0.

The x-coordinate decreases from 0 to -80 and then increases to 0; from there it increases further to 80.

Thus, the coordinates of the first car, after this 270-degree rotation, are (80, 0).

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3 years ago

Can someone help me with this please. ASAP!!

zhuklara [117]

x= 10 and y= 2

10-2= 8 and 10+2= 12

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3 years ago

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the pa

konstantin123 [22]

Answer:

$y=2x-8$

Step-by-step explanation:

The given parametric equation is;

$x=6+ln(t),y=t^2+3$

<h3><u>BY ELIMINATING THE PARAMETER</u></h3>

To eliminate the parameter we make $t$ the subject in one equation and put it inside the other.

We make $t$ the subject in $x=6+ln(t)$ because it is easier.

$\Rightarrow x-6=ln(t)$

$\Rightarrow {e}^{x-6}=e^{ln(t)}$

$\Rightarrow {e}^{x-6}=t$

Or

$t={e}^{x-6}$

We now substitute this into $y=t^2+3$ .

This gives us;

$y=(e^{x-6})^2+3$ .

$\Rightarrow y=e^{2(x-6)}+3$ .

We have now eliminated the parameter.

The equation of the tangent at (6,4) is given by;

$y-y_1=m(x-x_1)$

where the gradient function is given by;

$\frac{dy}{dx}=2e^{2(x-6)}$

We substitute $x=6$ into the gradient function to obtain the gradient.

$\Rightarrow m=2e^{2(6-6)}$

$\Rightarrow m=2e^0$

$\Rightarrow m=2$

The equation of the tangent becomes

$y-4=2(x-6)$

We simplify to obtain

$y=2x-12+4$

$y=2x-8$

<h3><u>WITHOUT ELIMINATING THE PARAMETER</u></h3>

The given parametric equation is;

$x=6+ln(t),y=t^2+3$

For $x=6+ln(t)$

$\frac{dx}{dt}=\frac{1}{t}$

For $y=t^2+3$

$\frac{dy}{dt}=2t$

The slope is given by;

$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\frac{dy}{dx}=\frac{2t }{\frac{1}{t} }$

$\frac{dy}{dx}=2t^2$

At the point, (6,4), we plug in any of the values into the parametric equation and find the corresponding value for $t$ .

Notice that

When $x=6$ , $6=6+\ln(t)$

$6-6=\ln(t)$

$0=\ln(t)$

$e^0=e^\ln(t)$

$1=t$

when $y=4$ , $4=t^2+3$

$4-3=t^2$

$1=t^2$

$t=\pm1$

But the slope is the same when we plug in any of these values for t.

$\frac{dy}{dx}=2(\pm1)^2=2$

The equation of the tangent becomes

$y-4=2(x-6)$

We simplify to obtain

$y=2x-12+4$

$y=2x-8$

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3 years ago

Sally’s mother bought two dozen pieces of fruit at the market. Six of the pieces of fruit were apples. Express the ratio of the

Strike441 [17]

1:4 expresses the total number of apples to the total pieces of fruit purchased

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3 years ago

Here is a trapezium, work out its area

ololo11 [35]

The rectangular part is 30 cm^2 and the triangular part is 10 because the base is the top of the trapezium subtracted from the base of 10.

So the overall area is 30 cm^2

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2 years ago