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mezya [45]
3 years ago
8

WILL MARK BRAINLIEST

Mathematics
1 answer:
alexandr1967 [171]3 years ago
4 0

Answer:

Equation: 7x - 23 = 25 + x

Solution: x = 8

Step-by-step explanation:

7x - 23 = 25 + x

1. Subtract 25 from both sides:

7x - 23 - 25 = 25 - 25 + x

7x - 48 = x

2. Subtract 7x from both sides to get x on one side:

7x - 7x - 48 = x - 7x

- 48 = -6x

3. Divide both sides by -6 to get x by itself:

8 = x

8 is the solution

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What is being done to the variable in the equation negative 5M = -40
Nataly [62]

Step-by-step explanation: In this equation, it's important to understand that the number 5 is the coefficient on the variable M. This just means that -5 "times" M equals -40. So, whenever you see a number next to a variable, it means multiplication.

If we were to solve, we would just divide both sides of the equation by 5. On the left the 5's cancel and on the right, -40 divided by 5 is -8.

8 0
4 years ago
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balu736 [363]

Answer:

B.) Associative Property of Addition

Step-by-step explanation:

This is because none of the numbers change places the only things that move are the parenthesis which correspond with the associative property.

4 0
3 years ago
What is the rule for 2.3, 3.2, 4.1, 5.0
AlladinOne [14]
It goes by order? :'l
6 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
Given these what is X if B-X=A
ser-zykov [4K]

Answer:

B

Step-by-step explanation:

B-X=A

or

B-A=X

or

X=B-A

X=\left[\begin{array}{ccc}-5&-1&6\\4&1&2\\0&-3&2\end{array}\right] -\left[\begin{array}{ccc}-1&-2&3\\4&8&-6\\0&1&5\end{array}\right] \\=\left[\begin{array}{ccc}-5+1&-1+2&6-3\\4-4&1-8&2+6\\0-0&-3-1&2-5\end{array}\right] \\=\left[\begin{array}{ccc}-4&1&3\\0&-7&8\\0&-4&-3\end{array}\right]

3 0
4 years ago
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