What are the coordinates of the terminal point determined by T=20pi/3

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pshichka [43]

I believe it is the second one hope that helped

4 0

2 years ago

Read 2 more answers

How many eighths are there in six and three quarters

Alex787 [66]

That questions has to do with equivalent fractions, express a fraction in eights.
six and three quarters
= 6 3/4
= 6 + 3/4
= 24/4 + 3/4
= 27/4
and to convert that fraction to eights, just multiply numerator and denominator by 2:
= (27*2)/(4*2)
= 54/8
there are 54 eights in six and three quarters

3 0

3 years ago

Each day her goal is to make 4 more baskets than she made thr day before. If she makes 10 baskets the first day and meets her go

irina [24]

Answer:

Janelle will make 62 baskets on 14th day.

Step-by-step explanation:

Given,

Number of goals on 1st day = 10

We need to find the number of goals on 14th day.

Solution,

Since Janelle made 10 goals on 1st day and want to make 4 more basket from the day before.

So we can say that;

$a_1=10$ and $d=4$

And also $n=14$

here $a_1$ = first term

d = common difference

n = number of terms

Now we apply the formula of A.P.

$T_n=a_1+(n-1)d$

now substituting the given values, we get;

$T_{14}=10+(14-1)4\\\\T_{14}=10+13\times 4\\\\T_{14}=10+52 =62$

Hence Janelle will make 62 basket on 14th day.

7 0

3 years ago

Which equation has a constant of proportionality equal to 1/2

Olin [163]

The equation is Oy = 3/6x

5 0

3 years ago

The coordinates of a particle in the metric xy-plane are differentiable functions of time t with:

frez [133]

Answer:

The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$ .

Step-by-step explanation:

This is an example of a related rate problem. A related rate problem is a problem in which we know one of the rates of change at a given instant $\frac{dx}{dt}$ and we want to find the other rate $\frac{dy}{dt}$ at that instant.

We know the rate of change of x-coordinate and y-coordinate:

$\frac{dx}{dt}=-2\:\frac{m}{s} \\\\\frac{dy}{dt}=-8\:\frac{m}{s}$

We want to find the rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12.

The distance of a point (x, y) and the origin is calculated by:

$s=\sqrt{x^2+y^2}$

We need to use the concept of implicit differentiation, we differentiate each side of an equation with two variables by treating one of the variables as a function of the other.

If we apply implicit differentiation in the formula of the distance we get

$s=\sqrt{x^2+y^2}\\\\s^2=x^2+y^2\\\\2s\frac{ds}{dt}= 2x\frac{dx}{dt}+2y\frac{dy}{dt}\\\\\frac{ds}{dt}=\frac{1}{s}(x\frac{dx}{dt}+y\frac{dy}{dt})$

Substituting the values we know into the above formula

$s=\sqrt{9^2+12^2}=15$

$\frac{ds}{dt}=\frac{1}{15}(9(-2)+12(-8))\\\\\frac{ds}{dt}=\frac{1}{15}\left(-18-96\right)\\\\\frac{ds}{dt}=\frac{1}{15}(-114)=-\frac{38}{5}=-7.6\:\frac{m}{s}$

The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$

7 0

3 years ago