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harkovskaia [24]
3 years ago
5

What is the image of (4, 3) after a dilation by a scale factor of 2 centered at the origin?​

Mathematics
1 answer:
emmasim [6.3K]3 years ago
5 0
I believe it’s (8,6) because dilation mean to make it bigger by 2
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Plz don’t be rude if i got brainly i’m gonna use it
SSSSS [86.1K]

Answer:

y=-5x-36

Step-by-step explanation:

-66-(-61)= -5

6-5= 1

-5/1= -5

Y=-5x + B

-66= -5(6) + B

-66= -30 + B

-66 + 30 = (-30 +30) + B

-36 = B

Y= -5x-36

4 0
4 years ago
(8.5-2x)(11-2x)(x) what is the approximate value of x that would allow you to construct an
Wittaler [7]

The largest volume possible from one piece of paper for open-top box is 64.296 cubic unit.

<h3>What is meant by the term maxima?</h3>
  • The maxima point on the curve will be the highest point within the given range, and the minima point will be the lowest point just on curve.
  • Extrema is the product of maxima and minima.

For the given question dimensions of open-top box;

The volume is given by the equation;

V = (8.5-2x)(11-2x)(x)

Simplifying the equation;

V = x(4x² - 39x + 93.5)

Differentiate the equation with respect to x using the product rule.

dV/dx = x(8x -39) + (4x² - 39x + 93.5)

dV/dx = 8x² - 39x + 4x² - 39x + 93.5

dV/dx = 12x² - 72x + 93.5

Put the Derivative equals zero to get the critical point.

12x² - 72x + 93.5 = 0.

Solve using quadratic formula to get the values.

x = 4.1  and x = 1.9

Put each value of x in the volume to get the maximum volume;

V(4.1) =  4.1(4(4.1)² - 39(4.1) + 93.5)

V(4.1) = 3.44 cubic unit.

V(1.9) = 1.9(4(1.9)² - 39(1.9) + 93.5)

V(1.9) = 64.296 cubic unit. (largest volume)

Thus, the largest/maximum volume possible from one piece of paper for open-top box is 64.296 cubic unit.

To know more about the maxima, here

brainly.com/question/17184631

#SPJ1

4 0
1 year ago
Rate of change between (2,11) and (8,14)
SCORPION-xisa [38]

Answer:

1/2

Step-by-step explanation:

the rate of change is = the change in y/ the change in x

if the points are (2, 11) and (8, 14)  

Rate of change = (14-11) / (8-2) = 3 / 6 = 1/2

8 0
3 years ago
A sample of gold has a mass of 579 g. The volume of the sample is 30 cm3. What is the density of the gold sample?
Ugo [173]

The density of the gold sample is 19.3g/cm³

Let the mass of the gold sample be represented by m

m = 579 g

Let the volume of the gold sample be represented by V

V = 30 cm³

Let the density of the gold sample be represented by ρ

The formula for the density is:

Density = \frac{Mass}{Volume}

\rho = \frac{m}{V} \\\rho = \frac{579}{30} \\\rho = 19.3 g/cm^3

The density of the gold sample is 19.3g/cm³

Learn more here: brainly.com/question/17780219

3 0
3 years ago
Which graph has a coefficient correlation r closeest to -0.95
V125BC [204]
Graph b has a correlation closest to -0.95
6 0
4 years ago
Read 2 more answers
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