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ki77a [65]
3 years ago
11

4. What is the area of a square with side lengths of 3/5 units?

Mathematics
1 answer:
irga5000 [103]3 years ago
4 0

Answer:

9/5 units²

Step-by-step explanation:

Area of a square=s²

A=(3/5)²

A=9/25

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kondaur [170]

Answer: D. Odd function

Step-by-step explanation:

x, -f(x) = f(-x)

7 0
1 year ago
15:45 in simplest form
mylen [45]
15:45 in simplest form is 1:3

the LCM of 15 and 45 is 15, and when you divide both by 15, you get 1:3
7 0
3 years ago
How do i graph the lines when the equations aren’t in slope intercept form????
jekas [21]

Slope-Intercept Form: y=mx+b

Standard Form: ax+by=c

Point- Slope: (y-y1)= m(x-x1)

There are multiple answers to your question-

  1. If you are only missing b(the y-intercept) but are given a set of points, plug the points into x and y and solve for b.
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7 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
What is the perimeter of this triangle? 15 cm 5 cm 25 cm 10 cm
Lady_Fox [76]

Answer:

55

Step-by-step explanation:

Add 15+5+25+10 which is 55

hope this helps! :)

3 0
2 years ago
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