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Veronika [31]
4 years ago
5

Which are perfect squares? Check all that apply.

Mathematics
2 answers:
alexandr1967 [171]4 years ago
5 0
36,16,81 64 has to be the answer hope it helps
Alja [10]4 years ago
4 0
Here they are:


36,16,81,64
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Please solve with explanation
Bas_tet [7]

Answer:

I don't know

Explanation:

Sorry :(

4 0
3 years ago
4. In the adjoining figure, ABCD is a
Marizza181 [45]

The missing figure is attached

(i) m∠APB = 90° ⇒ proved

(ii) AD = DP and PB = PC = BC ⇒ proved

(iii)  DC = 2 AD ⇒ proved

Step-by-step explanation:

ABCD is a  parallelogram in which:

  • m∠A = 60°
  • The bisectors of ∠A and ∠B meet DC at P

In parallelogram ABCD:

∵ m∠A = m∠C ⇒ opposite angles

∵ m∠A = 60°

∴ m∠C = 60°

∵ m∠A + m∠B = 180 ⇒ two adjacent supplementary angles

∴ 60 + m∠B = 180 ⇒ subtract 60 from both sides

∴ m∠B = 120°

∵ m∠B = m∠D ⇒ opposite angles

∴ m∠D = 120°

∵ AP is the bisector of angle A

- That means AP divide ∠A into two equal parts

∴ m∠BAP = m∠DAP = \frac{1}{2} m∠A

∴ m∠BAP = m∠DAP = \frac{1}{2} (60°)

∴ m∠BAP = m∠DAP = 30°

∵ BP is the bisector of angle B

- That means BP divide ∠B into two equal parts

∴ m∠ABP = m∠CBP = \frac{1}{2} m∠B

∴ m∠ABP = m∠CBP = \frac{1}{2} (120°)

∴ m∠ABP = m∠CBP = 60°

(i)

In ΔAPB

∵ m∠BAP = 30° ⇒ proved

∵ m∠ABP = 60° ⇒ proved

- Sum of the measures of the interior angles in a Δ is 180°

∴ m∠APB + m∠BAP + m∠ABP = 180°

∴ m∠APB + 30 + 60 = 180

- Add like terms in the left hand side

∴ m∠APB + 90 = 180

- Subtract 90 from both sides

∴ m∠APB = 90°

(ii)

In Δ ADP:

∵ m∠D = 120° ⇒ Proved

∵ m∠DAP = 30° ⇒ proved

- Sum of the measures of the interior angles in a Δ is 180°

∴ m∠APD + m∠DAP + m∠D = 180°

∴ m∠APD + 30 + 120 = 180

- Add like terms in the left hand side

∴ m∠APD + 150 = 180

- Subtract 150 from both sides

∴ m∠APD = 30°

∵ m∠DAP = m∠APD = 30°

- If two angles in a triangle are equal in measures, then the triangle

  is isosceles

∴ Δ ADP is an isosceles triangle

∴ AD = DP

In Δ BPC:

∵ m∠PBC = 60° ⇒ proved

∵ m∠C = 60° ⇒ proved

- Sum of the measures of the interior angles in a Δ is 180°

∴ m∠BPC + m∠PBC + m∠C = 180°

∴ m∠BPC + 60 + 60 = 180

- Add like terms in the left hand side

∴ m∠BPC + 120 = 180

- Subtract 120 from both sides

∴ m∠BPC = 60°

∵ m∠PBC = m∠C = m∠BPC = 60°

- If the three angles of a triangle are equal in measure, then

  the triangle is equilateral

∴ Δ BPC is an equilateral triangle

∴ PB = PC = BC

(iii)

∵ AD = BC ⇒ opposite sides in parallelogram

∵ AD = DP ⇒ Proved

- Equate the two right hand sides of AD

∴ BC = DP

∵ BC = PC

- Equate the right hand sides of BC

∴ DP = PC

∵ DC = DP + PC

∵ DP = AD

∴ PC = AD

- Substitute DP by AD and PC by AD in CD

∴ CD = AD + AD

∴ DC = 2 AD

Learn more:

You can learn more about parallelogram in brainly.com/question/6779145

#LearnwithBrainly

6 0
4 years ago
Calculate the volume of the solid. Round answer to the nearest hundredth.
LekaFEV [45]

Answer:

201.06

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
What are the values of x and y in the matrix equation below?
makvit [3.9K]
-8x-12+y^2
This is my aswer haha
3 0
4 years ago
Which statement is true about figures ABCD and A'B'C'D?
Anika [276]

Answer:

Step-by-step explanation:

To understand the transformation better we will take a point B first.

Coordinates of B → (3, 0)

When we rotate this point by 180° counterclockwise about the origin, rule to be followed,

(x, y) → (-x, -y)

By this rule, coordinates of B'' will be,

B(3, 0) → B''(-3, 0)

Now we translate the point B by 2 units right.

Rule for the translation is,

(x, y) → [(x + 2), 0]

Following this rule,

B''(-3, 0) → B'[(-3+2), 0]

             → B'(-1, 0)

We can conclude with the statement, "Figure ABCD was rotated by 180° about the origin followed by the translation of 2 units to the right" to get the new figure A'B'C'D' similar to pre-image ABCD.

5 0
3 years ago
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