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Sav [38]
3 years ago
13

Write five numbers that are eight digits long under expanded form

Mathematics
2 answers:
almond37 [142]3 years ago
6 0
12345678 2345678 3456789 9876543
Dmitriy789 [7]3 years ago
3 0
12345678, 87654321, 98765432, 23456789, 54789021
These are random numbers but are possible answers to the question
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many ® Black pencils cost N75 each and coloured pencils cost N105 each. If 24 mixed pencils cost #2010, how of them were black?
laila [671]

Answer:

85

Step-by-step explanation:

I hope my answer help you

7 0
3 years ago
6(1 − 3d) < 2(7d + 6
vagabundo [1.1K]

Answer:

d> -3/16

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
PLEASE HELPPP!!!!!!!
Rudik [331]
The answer is 11 because if you look at the chart. You can see everything can be divided by 8 and 88 divided by 8 is 11.
3 0
3 years ago
If ABCD is an A4 sheet and BCPO is the square, prove that △OCD is an isosceles triangle. And find the angles marked as 1 to 8 wi
Dmitry [639]

Answer:

The diagram for the question is missing, but I found an appropriate diagram fo the question:

Proof:

since OC = CD = 297mm Therefore, Δ OCD is an isoscless triangle

∠BCO = 45°

∠BOC = 45°

∠PCO = 45°

∠POC = 45°

∠DOP = 22.5°

∠PDO = 67.5°

∠ADO = 22.5°

∠AOD = 67.5°

Step-by-step explanation:

Given:

AB = CD = 297 mm

AD = BC = 210 mm

BCPO is a square

∴ BC = OP = CP = OB = 210mm

Solving for OC

OCB is a right anlgled triangle

using Pythagoras theorem

(Hypotenuse)² = Sum of square of the other two sides

(OC)² = (OB)² + (BC)²

(OC)² = 210² + 210²

(OC)² = 44100 + 44100

OC = √(88200

OC = 296.98 = 297

OC = 297mm

An isosceless tringle is a triangle that has two equal sides

Therefore for △OCD

CD = OC = 297mm; Hence, △OCD is an isosceless triangle.

The marked angles are not given in the diagram, but I am assuming it is all the angles other than the 90° angles

Since BC = OB = 210mm

∠BCO = ∠BOC

since sum of angles in a triangle = 180°

∠BCO + ∠BOC + 90 = 180

(∠BCO + ∠BOC) = 180 - 90

(∠BCO + ∠BOC) = 90°

since ∠BCO = ∠BOC

∴  ∠BCO = ∠BOC = 90/2 = 45

∴ ∠BCO = 45°

∠BOC = 45°

∠PCO = 45°

∠POC = 45°

For ΔOPD

Tan\ \theta = \frac{opposite}{adjacent}\\ Tan\ (\angle DOP) = \frac{87}{210} \\(\angle DOP) = Tan^-1(0.414)\\(\angle DOP) = 22.5 ^{\circ}

Note that DP = 297 - 210 = 87mm

∠PDO + ∠DOP + 90 = 180

∠PDO + 22.5 + 90 = 180

∠PDO = 180 - 90 - 22.5

∠PDO = 67.5°

∠ADO = 22.5° (alternate to ∠DOP)

∠AOD = 67.5° (Alternate to ∠PDO)

3 0
3 years ago
PLEASE ANSWER ALL BIG REWARD
sleet_krkn [62]

Answer:

  1. multiplying
  2. similar figures
  3. scale factor
  4. (x, y) (–3x, –3y) and (x, y) (0.23x, 0.23y)
  5. (x, y) --> (2x, 4y)
  6. Need diagram
  7. A. The line segment has become longer with endpoints D' (-12, -10) and E' (6 -8).
  8. Need image
  9. Reduction
  10. Enlargement

<em>good luck, i hope this helps :)</em>

6 0
3 years ago
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