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Aleks [24]
3 years ago
5

HELP ASAP HURRY HURRY

Mathematics
2 answers:
mojhsa [17]3 years ago
3 0
GIRL UHH TRY USING SOCRATIC HOLD ON
blsea [12.9K]3 years ago
3 0

Answer:

reflection across x=1 if I'm wrong sorry I cant see the screen well

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QUESTION 1: TRANSPORT Mr Dlamini transport people between Butterworth and East London using a bus which has a capacity of 100 pe
Dvinal [7]

From the series, it can be deduced that the sum of the first two terms will be 18.

<h3>How to calculate the series</h3>

From the information given, it was stated that the transport charge starts with a minimum charge of R8 and thereafter it was increased by R2 for each kilometre.

The series is also given as S. = 8 + 10 + 12 + 14 +... Therefore, the sum of the first two terms will be:

= 8 + 10

= 18

Expressing T2 in terms of S, and S2 will be T2 = S + S2.

Learn more about arithmetic series on:

brainly.com/question/24295771

5 0
2 years ago
A crawling tractor sprinkler is located as pictured below, 100 feet South of a sidewalk. Once the water is turned on, the sprink
deff fn [24]

Answer:

  a) see below

  b) 32 minutes after turn-on

  c) 52 minutes after turn-on

  d) 20 minutes

  e) 6856.6 ft²

Step-by-step explanation:

a) We have elected to put the origin at the point where the hose crosses the south edge of the sidewalk. Units are feet. Then the sprinkler starts at (0, -100). After 1 hour, 3600 seconds, the sprinkler is 1800 inches, or 150 ft north of where it started, so stops at (0, 50).

The lines forming the sidewalk boundaries are y=0 and y=10.

__

b) Water will first strike the sidewalk when the sprinkler is 20 feet south of it, or 80 feet north of where it started. The sprinkler travels that distance in ...

  (80 ft)(12 in/ft)/(1/2 in/s)(1 min/(60 s)) = 32 min . . . time to start sprinkling sidewalk

__

c) The sprinkler has to travel to a point 130 ft north of its starting position for the water to fall north of the sidewalk. That distance is traveled in ...

  (130 ft)(2/5 min/ft) = 52 min . . . time until end of sprinkling sidewalk

Note that we have combined the scale factors in the expression of part b into one scale factor of (2/5 min/ft).

__

d) The difference of times in parts b and c is the time water falls on the sidewalk: 20 minutes.

__

e) In one hour, the sprinkler travels a distance of ...

  (60 min)(5/2 ft/min) = 150 ft

Of that distance, 10 feet is sidewalk. So, the sprinkler covers an area of grass that is a 140 ft by 40 ft rectangle and a circle of 20 ft radius. The total area of that is ...

  A = LW + πr² = (140 ft)(40 ft) +π(20 ft)² = (14+π)(400) ft² ≈ 6856.6 ft²

The area of grass watered in 1 hour is about 6856.6 ft².

5 0
4 years ago
Which graphed line represents an equation of a line with a slope of -1 and a point on the line of (1, - 1)? A) red line
mamaluj [8]

Answer:

y = -x

Step-by-step explanation:

slope = change in y/change in x

slope = -1

Find the y-intercept

x value difference between (0,0) and (1,-1) = 1

if change in x = 1 then change in y = -1 so y-intercept = -1 + 1 = 0

y = -x + 0

4 0
3 years ago
What’s 2 plus 2 plus four plus eight plus six
Anastaziya [24]

Answer:

22

Step-by-step explanation:

2+2+4+8+6=22

5 0
3 years ago
Read 2 more answers
Pls someone help me with these 2 questions and explain.. ​
garik1379 [7]

question 6,

Answer : 35.7 degree from north

Step-by-step explanation:

Bearing is angle w.r.t the north.

To calculate this angle we need the angle of the triangle in the diagram.

We know two sides of the triangle the base (18) and Perpendicular(25).

Tan(x) = 25/18

arctan(25/18) = 54.24 deg

Subtracting from 90 to get angle w.r.t north 90-54.24 = 35.7

Question 7,

Solution: 33.12

Explaination

calculate the length of the straw within the glass.

It makes a right angle  triangle with the glass, where the length is the hypotenuse.

the base is 9 (radius *2 = 4.5*2)

using Cos(x) = base / hypotenuse

cos(72) = 9 /Hypotenuse

Hypotenuse = 9/cos(72)

Hypotenuse  = 29.12

total length is hypotenuse + the length protruding out(4)

29.12 + 4  = 33.12

6 0
3 years ago
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