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gavmur [86]
4 years ago
6

To add matrices add, ______________ entries. A) corresponding B) alternating

Mathematics
1 answer:
brilliants [131]4 years ago
8 0
A) corresponding
this is the answer hope it helps
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A lotion is made from an oil blend costing $1.50 per
Natalija [7]

Answer: (4-g) 1.5

Step-by-step explanation:

7 0
3 years ago
Would -x - 3y = -1 convert into y = -1/3 x - 1/3​
vfiekz [6]

Answer:

Yes, -x - 3y = -1 is the same as y = -1/3x - 1/3

Step-by-step explanation:

-x - 3y = -1

Add x to both sides which gives you

-3y = x-1

Divide both sides by -3 which gives you

y = -1/3x-1/3

Reminder whatever you do to one side you have to do to the other side!

Hope this helps!

3 0
3 years ago
Si n es un número impar, ¿cuál de las siguientes opciones representa un número par? a) 2n +1 b) n (n + 2) c) n + (n-1) e) 2 (n +
emmainna [20.7K]

Answer:

Option E) is correct

If n is an odd number then 2(n+1) represents an even number.

Step-by-step explanation:

Let X=set of all odd numbers

that is X={\{1,3,5,7...}\}

Let Y=set of all even numbers

that is Y={\{2,4,6,8...}\}

Verify that 2(n+1) represents an even number where n is an odd number:

Put n=1 in 2(n+1) we get

2(1+1)=2(2)=4

Put n=3 in 2(n+1) we get

2(3+1)=2(4)=8

Put n=5 in 2(n+1) we get

2(5+1)=2(6)=12

and so on.

From above results we get 2(n+1) represents an even number and is belongs to the set Y when n is an odd number.

Therefore Option E) is correct

If n is an odd number then 2(n+1) represents an even number.

5 0
3 years ago
Which statement is true?
EastWind [94]

Answer:

the last option

Step-by-step explanation:

not exactly sure how to explain it :/

7 0
3 years ago
Read 2 more answers
Does 23^-1 (mod 1000) exist? If yes solve it.
sweet [91]

Yes, 23 has an inverse mod 1000 because gcd(23, 1000) = 1 (i.e. they are coprime).

Let <em>x</em> be the inverse. Then <em>x</em> is such that

23<em>x</em> ≡ 1 (mod 1000)

Use the Euclidean algorithm to solve for <em>x</em> :

1000 = 43×23 + 11

23 = 2×11 + 1

→   1 ≡ 23 - 2×11   (mod 1000)

→   1 ≡ 23 - 2×(1000 - 43×23)   (mod 1000)

→   1 ≡ 23 - 2×1000 + 86×23   (mod 1000)

→   1 ≡ 87×23 - 2×1000 ≡ 87×23   (mod 1000)

→   23⁻¹ ≡ 87   (mod 1000)

3 0
3 years ago
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