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2 years ago

How do u square a scale factor?

Mathematics

2 answers:

seraphim [82]2 years ago

8 0

Answer:

The scale factor for area is found by squaring the scale factor for length. Thus the scale factor for area is s2. If this is reversed, a method is provided for solving the problem being investigated. The scale factor for area is 2.

Step-by-step explanation:

Dennis_Churaev [7]2 years ago

5 0

You multiply the area by the scale factor twice. Here is an example: if we have a rectangle that has a length 3 and a height of 4 and the scale drawing with a scale factor of 2, how many times bigger is the scale drawings area? The original shape is 3 by 4 so we multiply those to find the area of 12 square units.

hope it helps!!!

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The lengths of the sides of a triangle are represented by 3 consecutive even integers. if the perimeter of the triangle is 72 fe

iris [78.8K]

P = a + b + c
P = 72
a = x
b = x + 2
c = x + 4

72 = x + x + 2 + x + 4
72 = 3x + 6
72 - 6 = 3x
66 = 3x
66/3 = x
22 = x

x + 2 = 22 + 2 = 24
x + 4 = 22 + 4 = 26

the sides of ur triangle are : 22,24,26 <==

6 0

3 years ago

20k + 5 = 5k + 65 know the answer

Katyanochek1 [597]

Solving for 'k'

20k + 5 = 5k + 65

20k = 5k + 65 -5

20k = 5k + 60

20k - 5k = 60

15k = 60

15k/15 = 60/15

k = 4

4 0

3 years ago

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Which expression is equivalent to One-fifth (150 x minus 80 y + 50 minus 50 x minus 25 y + 20)?

nekit [7.7K]

Answer:

150 x - 80 y + 50 - 50 x - 25 y + 20

Arranging like terms 

150 x - 50 x -25 y - 80y + 50 + 20

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2 years ago

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Item 12

zlopas [31]

Answer: about 14pi inches

Explanation:

7 * 6.28319 = 43.9823

43.9823 / 3.14159 = 14.0000127324

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3 years ago

Verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given poi

lara [203]

For each curve, plug in the given point $(x,y)$ and check if the equality holds. For example:

(I) (2, 3) does lie on $x^2+xy-y^2=1$ since 2^2 + 2*3 - 3^2 = 4 + 6 - 9 = 1.

For part (a), compute the derivative $\frac{\mathrm dy}{\mathrm dx}$ , and evaluate it for the given point $(x,y)$ . This is the slope of the tangent line at the point. For example:

(I) The derivative is

$x^2+xy-y^2=1\overset{\frac{\mathrm d}{\mathrm dx}}{\implies}2x+x\dfrac{\mathrm dy}{\mathrm dx}+y-2y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x+y}{2y-x}$

so the slope of the tangent at (2, 3) is

$\dfrac{\mathrm dy}{\mathrm dx}(2,3)=\dfrac74$

and its equation is then

$y-3=\dfrac74(x-2)\implies y=\dfrac74x-\dfrac12$

For part (b), recall that normal lines are perpendicular to tangent lines, so their slopes are negative reciprocals of the slopes of the tangents, $-\frac1{\frac{\mathrm dy}{\mathrm dx}}$ . For example:

(I) The tangent has slope 7/4, so the normal has slope -4/7. Then the normal line has equation

$y-3=-\dfrac47(x-2)\implies y=-\dfrac47x+\dfrac{29}7$

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3 years ago