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Talja [164]
3 years ago
11

Using the exchange rate £1 = $1.34, calculate how many £ are in $7.25. Give your answer rounded to 2 dp.

Mathematics
2 answers:
ipn [44]3 years ago
6 0

Answer:

£5.41

Step-by-step explanation:

7.25/1.34 = 5.41044...

Round = 5.41

Anettt [7]3 years ago
4 0
Pineapple galaxy is right!
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Find the midpoint of the line segment JK if j (4,6) and k(0,-4)
Snowcat [4.5K]

Answer:

(2,1)

General Formulas and Concepts:

Order of Operations: BPEMDAS

Midpoint Formula: (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

Step-by-step explanation:

<u>Step 1: Define points</u>

J (4, 6)

K (0, -4)

<u>Step 2: Find midpoint</u>

  1. Substitute:                         (\frac{4+0}{2},\frac{6-4}{2})
  2. Add/Subtract:                    (\frac{4}{2},\frac{2}{2})
  3. Divide:                               (2,1)
4 0
3 years ago
In the diagram shown, line m is parallel to line n. Robert says that ∠2 and ∠7 are supplementary angles. Is he correct? Why or w
bekas [8.4K]

Robert is not correct because in order for it to be supplementary the angles have to be consecutive interior

6 0
3 years ago
PLEASE HELP GIVING BRAINLIST
melisa1 [442]
I’m pretty sure it’s the number of wrenches in the toolbox because that gonna affect the amount of space he has left
3 0
2 years ago
Solve the equation <br> X+ (-15) = 9
ella [17]

Step-by-step explanation:

x + ( - 15) = 9

x - 15 = 9

x = 9  + 15

x = 24

3 0
3 years ago
Read 2 more answers
How to work it out logically? Question a !!!
Arturiano [62]

Answer:

Fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.

Step-by-step explanation:

Given the figure with dimensions. we have to find the area of given figure.

Area of figure=ar(1)+ar(2)+ar(3)

Area of region 1 = ar(ANGI)+ar(AIB)

                          =L\times B+\frac{1}{2}\times base\times height\\\\=[1500\times (5000-2000-1500)]+\frac{1}{2}\times (3000-1500)\times (5000-2000-1500)\\\\=3375000m^2=337.5ha

Area of region 2 = ar(DHBC)

                       =2000\times1500\\\\=3000000m^2=300ha

Area of region 3 = ar(GFEH)

                             (2000+1500)\times 1000\\\\=3500000m^2=350ha

Hence, Area of figure=ar(1)+ar(2)+ar(3)=337.5ha+300ha+350ha

                                                 =987.5 ha

Now, we have to do straight-line fencing such that area become half and cost of fencing is minimum.

Let the fencing be done through x m downward from B which divides the two into equal area.

⇒ Area of upper part above fencing=Area of lower part below fencing

⇒ar(ANGB)+ar(GKLB)=ar(KLCM)+ar(MDCF)\\\\337500+3000x=(3500-x)\times 1000+2000(1500-x)\\\\3375000+3000x=3500000-1000x+3000000-2000x\\\\6000x=315000\\\\x=\frac{315000}{6000}=520.8m

Hence, fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.

7 0
3 years ago
Read 2 more answers
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