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3 years ago

Point Q lies on the circle and has an x-coordinate of 4.

Mathematics

2 answers:

Shtirlitz [24]3 years ago

5 0

Answer:

Step-by-step explanation:

Ed2020

spin [16.1K]3 years ago

4 0

Answer:

2 square root 5

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

The equation of the circle is equal to

$(x-h)^2+(y-k)^2=r^2$

where

(h,k) is the center and r is the radius of the circle

In this problem we have

center at (0,0) and radius equal 6 units

$x^2+y^2=6^2\\x^2+y^2=36$

Remember that

If a point lie on the circle, then the point must satisfy the equation of the circle

Substitute the x-coordinate of point Q in the equation of the circle and solve for the y-coordinate of point Q

For x=4

$4^2+y^2=36\\y^2=36-16\\y^2=20$

$y=\pm\sqrt{20}$

simplify

$y=\pm2\sqrt{5}$

The point Q lie on the first Quadrant (see the picture)

The y-coordinate is positive

therefore

The y-coordinate of point Q is $2\sqrt{5}$

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7x − 5 < 13x − 3 <br><br>PLz anyone

Ksju [112]

7x-5<13x-3

Minus 7x

-5<6x-3

then add 3

-2<6x

Then divide -2 by 6 and 6x by 6

-1/3>x

I think thats right

4 0

3 years ago

Find the center of mass of the wire that lies along the curve r and has density =4(1 sin4tcos4t)

dolphi86 [110]

The mass of the wire is found to be 40π√2 units.

To calculate the mass of the wire which runs along the curve r ( t ) with the density function δ=5.

The general formula is,

Mass = $\int_a^b \delta\left|r^{\prime}(t)\right| d t$

To find, we must differentiate this same given curve r ( t ) with respect to t to estimate |r'(t)|.

The given integration limits in this case are a = 0, b = 2π.

Now, as per the question;

The equation of the curve is given as;

r(t) = (4cost)i + (4sint)j + 4tk

Now, differentiate this same given curve r ( t ) with respect to t.

$\begin{aligned}\left|r^{\prime}(t)\right| &=\sqrt{(-4 \sin t)^2+(4 \cos t)^2+4^2} \\&=\sqrt{16 \sin ^2 t+16 \cos ^2 t+16} \\&=\sqrt{16\left(\sin t^2+\cos ^2 t\right)+16}\end{aligned}$

Further simplifying;

$\begin{aligned}&=\sqrt{16(1)+16} \\&=\sqrt{16+16} \\&=\sqrt{32} \\\left|r^{\prime}(t)\right| &=4 \sqrt{2}\end{aligned}$

Now, use integration to find the mass of the wire;

$\begin{aligned}&=\int_a^b \delta\left|r^{\prime}(t)\right| d t \\&=\int_0^{2 \pi} 54 \sqrt{2} d t \\&=20 \sqrt{2} \int_0^{2 \pi} d t \\&=20 \sqrt{2}[t]_0^{2 \pi} \\&=20 \sqrt{2}[2 \pi-0] \\&=40 \pi \sqrt{2}\end{aligned}$