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Alja [10]
3 years ago
11

The double number line shows that Toni can type 180180180 words in 222 minutes. Based on the ratio shown in the double number li

ne, how many words can Toni type in 333 minutes?
Mathematics
2 answers:
vfiekz [6]3 years ago
6 0

Answer:

<h3>Toni can type words in 3 minutes is 270 words</h3>

Step-by-step explanation:

Given that the double number line shows that Toni can type 180 words in 2 minutes.

<h3>To find how many words can Toni type in 3 minutes:</h3>

From the given Toni can type 180 words in 2 minutes,

⇒ Toni can type \frac{180}{2}

= 90

<h3>∴ 90 words per minute </h3>

For Toni would be able to type words in 3 minutes is

90\times 3

= 270

<h3>∴ 270 words per minute.</h3><h3>∴ Toni can type words in 3 minutes is 270 words.</h3>
Anna007 [38]3 years ago
4 0

Answer:

put 270 and the anwser shoud be correct

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Factor the trinomial:<br> 3x2 + 13x + 12
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Step-by-step explanation:

(3x + 4)(x + 3)

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3x^2 + 13x + 12

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Determine if the slope of the line graphed below is either positive or negative. Then find the slope (m) of the line
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Given that the measurement is in centimeters, find the area of the circle to the nearest tenth. (Use 3.14 for π)
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Answer:

Option D) 254.3 cm2

Step-by-step explanation:

we know that

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A ladder 20 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2
alukav5142 [94]

Answer:

0.17 °/s

Step-by-step explanation:

Since the ladder is leaning against the wall and has a length, L and is at a distance, D from the wall. If θ is the angle between the ladder and the wall, then sinθ = D/L.

We differentiate the above expression with respect to time to have

dsinθ/dt = d(D/L)/dt

cosθdθ/dt = (1/L)dD/dt

dθ/dt = (1/Lcosθ)dD/dt where dD/dt = rate at which the ladder is being pulled away from the wall = 2 ft/s and dθ/dt = rate at which angle between wall and ladder is increasing.

We now find dθ/dt when D = 16 ft, dD/dt = + 2 ft/s, and L = 20 ft

We know from trigonometric ratios, sin²θ + cos²θ = 1. So, cosθ = √(1 - sin²θ) = √[1 - (D/L)²]

dθ/dt = (1/Lcosθ)dD/dt

dθ/dt = (1/L√[1 - (D/L)²])dD/dt

dθ/dt = (1/√[L² - D²])dD/dt

Substituting the values of the variables, we have

dθ/dt = (1/√[20² - 16²]) 2 ft/s

dθ/dt = (1/√[400 - 256]) 2 ft/s

dθ/dt = (1/√144) 2 ft/s

dθ/dt = (1/12) 2 ft/s

dθ/dt = 1/6 °/s

dθ/dt = 0.17 °/s

8 0
3 years ago
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