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Mrrafil [7]
3 years ago
8

Emma buys a house for £201 500

Mathematics
2 answers:
Novay_Z [31]3 years ago
4 0

Answer:

6%

Step-by-step explanation:

\frac{213590 - 201500}{201500} = \frac{12090}{201500} = 6

lisabon 2012 [21]3 years ago
3 0

Answer:

6%

Step-by-step explanation:

Original price = £201 500

Selling price = £213 590

The profit equals = Selling price - Original price

213590 - 201500 = 12090

The profit is £12090.

The profit percent equals = Profit/Original price × 100

12090/201500 × 100

6/100 × 100 = 6

The profit percentage was 6%

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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
Find the distance travelled in 1 h 40 min at a speed of 90 km/h
Mariulka [41]
90 km/h is your constant.
1h 40m - 1h since we have 90 km/h.
2/3 of 90 is what we need to find for the 40 min, since 40m/60m = 2/3.
60 is 2/3 of 90.
90 + 60 = 150.
Your distance traveled is 150km.
5 0
3 years ago
Find the slope-intercept equation of the line
Natali [406]

Answer:

y=8.

Step-by-step explanation:

all the details are in the attachment.

8 0
2 years ago
Which sequences of transformations confirm the congruence of shape II and shape I?
mylen [45]

Answer:

1. When we reflect the shape I along X axis it will take the shape I in first quadrant, and then if we rotate the shape I by 90° clockwise, it will take the shape again in second quadrant . So we are not getting shape II. This Option is Incorrect.

2. Second Option is correct , because by reflecting the shape I across  X axis and then by 90° counterclockwise rotation will take the Shape I in second quadrant ,where we are getting shape II.

3. a reflection of shape I across the y-axis followed by a 90° counterclockwise rotation about the origin takes the shape I in fourth Quadrant. →→ Incorrect option.

4. This option is correct, because after reflecting the shape through Y axis ,and then rotating the shape through an angle of 90° in clockwise direction takes it in second quadrant.

5. A reflection of shape I across the x-axis followed by a 180° rotation about the origin takes the shape I in third quadrant.→→Incorrect option

6 0
2 years ago
Read 2 more answers
If you were to pull colored marbles out of the bag (one at a time, and putting the marble back each time) for 500 tries, approxi
shepuryov [24]

Answer:

We need to know the other marbles in the bag and how many there are of each.

5 0
2 years ago
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