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aleksley [76]
2 years ago
12

What's the correct answer and why?

Mathematics
1 answer:
alexandr402 [8]2 years ago
4 0
3x^4 becomes 3(3x)^4=3*3^4*x^4
y^2 becomes (3y)^2=3^2y²
the new z=3² the old z, so b is correct.
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7. The probability that Jane smokes is 4/9. The probability
svet-max [94.6K]

To find the probability that she develops lung cancer given that she smokes, multiply the two probabilities together:

4/9 x 3/10 = (4 * 3) / (9*10) = 12/90 = 2/15

6 0
2 years ago
Set up, but do not evaluate, the integral that represents the length of the curve given by
netineya [11]

Answer:

L = ∫₀²ᵖⁱ √((1 − sin t)² + (1 − cos t)²) dt

Step-by-step explanation:

Arc length of a parametric curve is:

L = ∫ₐᵇ √((dx/dt)² + (dy/dt)²) dt

x = t + cos t, dx/dt = 1 − sin t

y = t − sin t, dy/dt = 1 − cos t

L = ∫₀²ᵖⁱ √((1 − sin t)² + (1 − cos t)²) dt

Or, if you wish to simplify:

L = ∫₀²ᵖⁱ √(1 − 2 sin t + sin²t + 1 − 2 cos t + cos²t) dt

L = ∫₀²ᵖⁱ √(3 − 2 sin t − 2 cos t) dt

7 0
3 years ago
PLEASE HELPPppp
sasho [114]

Answer: B

Hope this helps :)

5 0
2 years ago
a map has a scale of 1 : 50000. Bridgetown is 16.7km away from Littleton. What is the distance between these towns on the map?
lisov135 [29]

Answer:

if it is 1 cm answer is 3340cm

Step-by-step explanation:

1m = 100 cm

500m = 50000cm

change 16 .7 to m

16700

divide

16700/500

33.4

change to cm = 3340

3 0
3 years ago
The line of reflection will (always, sometimes, never) be the perpendicular bisector between a preimage and it’s reflected image
frez [133]

Answer:

Step-by-step explanation:

line of reflection: <em>a fixed line directly in the middle of the pre-image and the reflected image. </em>

perpendicular: <em>two lines meeting at a 90 degree angle</em>

bisector: <em>line that divides something into two equal parts</em>

pre-image: <em>the original figure before any transformation or translation has been made</em>

reflected image: <em>the original image after being reflected across the line of reflection.</em>

<em />

Is the line of reflection a line running perpendicular, directly in the middle of the pre-image and the reflected image? Yes, always would be my answer.

6 0
3 years ago
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