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3 years ago

Choose the following rule that CORRECTLY describes a dilation :

Mathematics

1 answer:

kkurt [141]3 years ago

4 0

I believe that answer choice D is correct, I know it is not C, A or D unless your teacher made a mistake presenting answer C. Hope this helps.

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A motarcar is running at the rate of 66 km per hour. Each other its wheels makes 500 revolutios per minute. find the radius of t

wolverine [178]

Answer:

simple

Step-by-step explanation:

The radius of the wheel can be solved from its given speed and RPM.

Converting km/h to m/min

speed = 66 km / h = 66000 m / 60 min

speed = 1100 m / min

Converting one revolution to meters

1100 m / min = 500 revolution / min ... (given)

500 revolutions = 1100 m

1 revolution = 2.2 m

Solving for the radius of the wheels in meters

1 revolution = 2.2 m

2 * pi * radius = 2.2 m

radius = 2.2 m / 2 * pi

radius = 1.1 m / pi

radius = 0.3501 m

3 0

3 years ago

Gavin made a wooden planter in the shape of a cube. He measured the inside of the planter and found that each side is 38.75 cm l

notka56 [123]

Answer:

3/5

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Can someone please help me

marta [7]

Answer:

when u divide both side by 2 the new equation will be y-5x=-3... and equation of a line is written in form of y=mx+c

so y=5x-3

3 0

3 years ago

Find the derivative of the function at P 0 in the direction of A. f(x,y,z) = 3 e^x cos(yz), P0 (0, 0, 0), A = - i + 2 j + 3k

Alik [6]

The derivative of $f(x,y,z)$ at a point $p_0=(x_0,y_0,z_0)$ in the direction of a vector $\vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k$ is

$\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}$

We have

$f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k$

and

$\vec a=-\vec\imath+2\,\vec\jmath+3\,\vec k\implies\|\vec a\|=\sqrt{(-1)^2+2^2+3^2}=\sqrt{14}$

Then the derivative at $p_0$ in the direction of $\vec a$ is

$3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}$

3 0

4 years ago

Prove that angle ADC is congruent to angle CBA <br><br> with solution!

vova2212 [387]

Step-by-step explanation:

< i'll use this too represent angle

this is only presuming that this is a parallelogram and i also don't know what those numbers mean

< ACB = < CAD (alt <s of DA II CB) A

< ACD = < CAB (alt <s of DC II AB) A

AB is common S

therefore, ∆ADC is congruent to ∆CBA

if ∆ADC is congruent to ∆CBA

then < ADC = < CBA

5 0

3 years ago