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3 years ago

je Jessie competes in a hot dog eating contest he can eat in 8 minutes how many can he eat in 12 minutes

Mathematics

2 answers:

WITCHER [35]3 years ago

5 0

He can eat about 3-4 hot dogs

Tems11 [23]3 years ago

4 0

How many u eat in eight mins divided by 2 then and it to the hot dogs u are in 8mins

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5) If Oskar bisects the diameter of a circle, what is he trying to construct?

pav-90 [236]

Answer:

Center = (5,-3) and radius = 7

Step-by-step explanation:

(5) If Oskar bisects the diameter of a circle, what is he trying to construct the radius of the circle.

(6) The given equation is :

$(x-5)^2+(y+3)^2=49$

The general equation of circle is given by :

$(x-a)^2+(y-b)^2=r^2$ ...(1)

Where

(a,b) are the coordinates of the centre and r is the radius.

The given equation can be written as :

$(x-5)^2+(y+3)^2=7^2$ ..(2)

Comparing equation (1) and (2) we get :

a = 5, b = -3 and r = 7

So,

Center = (5,-3) and radius = 7

Hence, this is the required solution.

8 0

2 years ago

A truck driver drove 48 miles for 45 minutes. At this rate, how many miles can the truck driver drive in one hour?

alexandr1967 [171]

Answers

2880

Step-by-step explanation:

48 miles = 1 minute

? = 60 minutes( 1hr )

1hr = 48 × 60 = 2880miles

7 0

2 years ago

Miguel is a cartoonist. Miguel drew 2 cartoon cells in 3 hours. If he keeps drawing at the same rate, about how many cells will

Sonja [21]

In three hours Miguel drew 2 cartoon cells . And since he keeps drawing at the same rate, therefore,

in one hour Miguel drew (2/3) cartoon cells .

Hence in 9 hours, Miguel drew (2/3)* 9 = 18/3=6 cartoon cells .

So the answer of the given question is 6 cartoon cells .

4 0

3 years ago

Read 2 more answers

interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm

Igoryamba

Answer:

The curvature is $\kappa=1$

The tangential component of acceleration is $a_{\boldsymbol{T}}=0$

The normal component of acceleration is $a_{\boldsymbol{N}}=1 (2)^2=4$

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}$

where

$\boldsymbol{T}}$ is the unit tangent vector.

$\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ is the speed of the object

We need to find $\boldsymbol{r}'(t)$ , we know that $\boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k}$ so

$\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}$

Next , we find the magnitude of derivative of the position vector

$|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2$

The unit tangent vector is defined by

$\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}$

$\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)$

We need to find the derivative of unit tangent vector

$\boldsymbol{T}'=\frac{d}{dt}(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})$

And the magnitude of the derivative of unit tangent vector is

$||\boldsymbol{T}'||=2\sqrt{\cos ^2\left(x\right)+\sin ^2\left(x\right)} =2$

The curvature is

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}=\frac{2}{2} =1$

The tangential component of acceleration is given by the formula

$a_{\boldsymbol{T}}=\frac{d^2s}{dt^2}$

We know that $\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ and $||\boldsymbol{r}'(t)}||=2$

$\frac{d}{dt}\left(2\right)\: = 0$ so

$a_{\boldsymbol{T}}=0$

The normal component of acceleration is given by the formula

$a_{\boldsymbol{N}}=\kappa (\frac{ds}{dt})^2$

We know that $\kappa=1$ and $\frac{ds}{dt}=2$ so

$a_{\boldsymbol{N}}=1 (2)^2=4$

3 0

3 years ago

Please help this is important for me.

Anit [1.1K]

<h3>Answer: point Q is located at (-1, 1)</h3>

=================================================

Explanation:

Check out the diagram below.

Plot R(-3,7) and T(3,-11) on the same xy grid.

Draw a vertical line through R and a horizontal line through T. A right triangle forms. At the intersection point is point S(-3,-11)

----------------------

Now measure the distance from point S to point T. You can count out the spaces or subtract the x coordinates and use absolute value.

|x1-x2| = |-3-3| = |-6| = 6

From point S to point T is 6 units.

We want to subdivide this horizontal length in the ratio 1:2

What this means is that we want to plot a point U somewhere such that SU:UT = 1:2

In other words,

SU = x

UT = 2x

SU+UT = ST

x+2x = 6

3x = 6

x = 6/3

x = 2

So we must move 2 spaces to the right from point S to get to U(-1,-11)

Going from point U(-1,-11) to T(3,-11) is 4 spaces

We have SU:UT = 2:4 = 1:2 to help confirm we have the correct location for point U

From point U, we then move straight up to the line segment RT

We'll land on Q(-1,1)

----------------------

Another way to find the y coordinate of point Q is to subdivide the segment RS into the ratio 1:2 similar to how we divided ST up

Segment RS is 18 units long since we go from y = 7 to y = -11 when going from R to S.

If V was on segment RS such that

RV:VS = 1:2

and RV = y

then

RV+VS = RS

y+2y = 18

3y = 18

y = 6

RV = y = 6

VS = 2y = 2*6 = 12

So you'll move 6 units down from y = 7 to land on y = 1 (when going from the y coordinate of R to the y coordinate of Q)

3 0

3 years ago