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Citrus2011 [14]
3 years ago
13

Pls pls help me with this

Mathematics
1 answer:
rosijanka [135]3 years ago
6 0

Answer:

Please check the explanation.

Step-by-step explanation:

                                                        6)

Given ΔMNO ≅ ΔPQR

As congruent parts of congruent triangle are congruent.

and also given

  • m∠N = 2x+6
  • m∠Q = 4x+12

so

m∠N = m∠Q

2x+6 = 4x+12

4x-2x = 6-12

2x  = -12

Divide both sides by 2

x = -6

Thus, the value of x = -6

Therefore, option 'A' is true.

                                                     7)

Given ΔABC ≅ ΔDEF

As congruent parts of congruent triangle are congruent.

and also given

  • AB = 4v+2
  • DE = 6v-4

so

AB = DE

4v+2 = 6v-4

6v-4v = 2+4

2v=6

Divide both sides by 2

v = 3

Thus, the value of v = 3

Therefore, option 'A' is true.

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B. the second one

Step-by-step explanation:

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What is the property of yx7=7y
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This is the commutative property of multiplication.

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This property states that we can multiply numbers in any order and still get the same number. Each side is the same operation in a different order.

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Which of the following functions is graphed below
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4 0
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Write the decimal 2.91 as a fraction or mixed number,
patriot [66]

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291/100

Step-by-step explanation:

3 0
3 years ago
Given the Arithmetic series A1+A2+A3+A4 7 + 11 + 15 + 19 + . . . + 91 What is the value of sum?
aivan3 [116]

Answer:

The value of the sum is 1078

Step-by-step explanation:

* Lets revise how to find the sum of the arithmetic series

- In the arithmetic series there is a constant difference between each

 two consecutive terms

- Ex:

# 4 , 9 , 14 , 19 , 24 , .......... (+ 5)

# 25 , 15 , 5 , -5 , -15 , .......... (-10)

- So if the first term is a and the constant difference between each two

  consecutive terms is d, then

  U1 = a , U2 = a + d , U3 = a + 2d , U4 = a + 3d , ..........

- Then the nth term is Un = a + (n - 1)d

- The sum of n terms is Sn = n/2[a + L] , where L is the last term in

 the series

* Lets solve the problem

∵ The arithmetic series is 7 + 11 + 15 + 19 + ......................... + 91

∴ The first term (a) is 7

∴ The last term is (L) 91

- Lets find the constant difference

∵ 11 - 7 = 4 and 15 - 11 = 4

∴ The constant difference (d) is 4

- Lets find the sum of the series from 7 to 91

∵ Sn = n/2[a + L]

∵ a = 7 and L = 91

∴ Sn = n/2[7 + 91]

- We need to know how many terms in the series

∵ L is the last term and equals 91 lets find its position in the series

∵ Un = a + (n - 1)d

∵ a = 7 , d = 4 and Un = 91

∴ 91 = 7 + (n - 1)(4) ⇒ subtract 7 from both sides

∴ 84 = (n - 1)(4) ⇒ divide both sides by 4

∴ 21 = n - 1 ⇒ add 1 to both sides

∴ n = 22

∴ The number of the terms in the series is 22

- Lets find the sum of the 22 terms (S22)

∴ S22 = 22/2[7 + 91]

∴S22 = 11[98] = 1078

* The value of the sum is 1078

4 0
3 years ago
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