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ioda
2 years ago
6

What are the coordinates of R' for the dilation D(0.5. p)(PQRS)?

Mathematics
1 answer:
RideAnS [48]2 years ago
6 0

Answer:

Step-by-step explanation:

Three step process.

Step 1, convert Q, R, P, and S to coordinates measured from point P rather than from the origin by subtracting P from each point:

Q' = Q-P = (-3-(-3),2-(-3)) = (0,5)

R' = R-P = (1-(-3),2-(-3)) =(4,5)

P' = P-P = (-3-(-3),-3-(-3)) =(0,0)

S' = S-P = (1-(-3),-3-(-3)) = (4,0)

Step 2, Apply the dilation factor (0.5):

Q'' = (0.5)Q' = (0.5*0, 0.5*5) = (0,2.5)

R'' = (0.5)R' = _____

P''= (0.5)P' = _____

S''= (0.5)S' = _____

Step 3, Convert coordinates back to the origin (vice point P) by adding P to each point.  These are the dilated points.

Q''' = Q'' + P = (0+(-3),2.5+(-3)) = (-3, -0.5)

R''' = R'' + P = _____

P''' = P'' + P = _____

S''' = S'' + P = _____

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noname [10]

The equation r=\sqrt\frac{V}{3.14h} can be used to find the radius.

Step-by-step explanation:

Given,

Volume of large can;

V=πr(r)h

V=πr²h

Dividing both sides by πh

\frac{V}{\pi h}=\frac{r^2\pi h}{\pi h}\\\\\frac{V}{\pi h}=r^2\\\\r^2=\frac{V}{\pi h}

Taking square root on both sides

\sqrt{r^2}=\sqrt{\frac{V}{\pi h}}\\r=\sqrt\frac{V}{\pi h}

Putting π=3.14

r=\sqrt\frac{V}{3.14h}

The equation r=\sqrt\frac{V}{3.14h} can be used to find the radius.

Keywords: volume, square root

Learn more about square root at:

  • brainly.com/question/10978510
  • brainly.com/question/11007026

#LearnwithBrainly

5 0
2 years ago
Eighty percent of one store's customers paid with credit cards. Forty customers came in that day. How many customers paid for th
mario62 [17]
80%=.8

40*.8=32

32 people made their purchases with credit cards.
6 0
2 years ago
Read 2 more answers
EXAMPLE 5 If F(x, y, z) = 4y2i + (8xy + 4e4z)j + 16ye4zk, find a function f such that ∇f = F. SOLUTION If there is such a functi
Valentin [98]

If there is such a scalar function <em>f</em>, then

\dfrac{\partial f}{\partial x}=4y^2

\dfrac{\partial f}{\partial y}=8xy+4e^{4z}

\dfrac{\partial f}{\partial z}=16ye^{4z}

Integrate both sides of the first equation with respect to <em>x</em> :

f(x,y,z)=4xy^2+g(y,z)

Differentiate both sides with respect to <em>y</em> :

\dfrac{\partial f}{\partial y}=8xy+4e^{4z}=8xy+\dfrac{\partial g}{\partial y}

\implies\dfrac{\partial g}{\partial y}=4e^{4z}

Integrate both sides with respect to <em>y</em> :

g(y,z)=4ye^{4z}+h(z)

Plug this into the equation above with <em>f</em> , then differentiate both sides with respect to <em>z</em> :

f(x,y,z)=4xy^2+4ye^{4z}+h(z)

\dfrac{\partial f}{\partial z}=16ye^{4z}=16ye^{4z}+\dfrac{\mathrm dh}{\mathrm dz}

\implies\dfrac{\mathrm dh}{\mathrm dz}=0

Integrate both sides with respect to <em>z</em> :

h(z)=C

So we end up with

\boxed{f(x,y,z)=4xy^2+4ye^{4z}+C}

7 0
1 year ago
Anyone know the answer??
olga_2 [115]
No. no i do not so yeah.
5 0
1 year ago
Which could be the graph of f(x) = |x - h| + k if h and k are both positive? On a coordinate plane, an absolute value graph has
Andrew [12]

Answer:

Step-by-step explanation:

f(x) = |x - h| + k has a vertex at (h, k), where both h and k are positive.  Only

"On a coordinate plane, an absolute value graph has a vertex at (2, 1)"  satisfies those requirements.

4 0
1 year ago
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