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Phoenix [80]
3 years ago
13

What is 33 photos into two grours so the ratio is 4 to 7

Mathematics
2 answers:
Keith_Richards [23]3 years ago
7 0
ratio:\\4:7\to4+7=11\\\\33:11=3\\\\therefore\\\\4:7\\\downarrow\\4\times3=12\\7\times3=21\\\\Answer\huge\boxed{12:21}
lakkis [162]3 years ago
6 0
You have 33 photos

You need to separate the photos in two groups for a ratio of 4:7

So what you do is add both numbers of the ratio

 4+7=11

Then divide the 33 photos by 11

33÷11=3

So that means that each number in the ratio is equivalent to 3 photos

Now multiply each ratio by 3

4*3=12 and 7*3=21

So the ratio of the photos are 12 to 21 or 12:21

Answer: ratio is 12 to 21 or can be written as 12:21
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-4(8r+3)

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Explain why a positive times a negative is a negative number.
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<u>2×(-3)</u>

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<u>Thus, a positive times a negative is a negative number.</u>

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Point A is located at (-3, 9) and point B is located at (12,-10). Find the distance from point A to point B rounded to the neare
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2 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)

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3 years ago
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