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katrin [286]
2 years ago
10

How do you make an equation have zero solutions

Mathematics
1 answer:
Anna35 [415]2 years ago
3 0
Well if the equation is undefined then that counts as having no solutions

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The top one I need help with
brilliants [131]
Same thing as what you did on the bottom. Find numbers with both 7 as the base and numbers that add to 14 on the top. Possibilities:
1) 7^10•7^4
2)7^6•7^8
37^2•7^12
5 0
3 years ago
a jewerly shop shop sells 240 necklaces in a month. 180 of the necklaces were sold via the shops, website, the rest were sold in
Hunter-Best [27]

Answer:

<h3>The workout ratio of the online sales to shop sales is 1:3.</h3>
7 0
2 years ago
A white rhinoceros weighs about 2 ½ tons. How many pounds does the white rhinoceros weigh?
steposvetlana [31]

Answer:

5000 pounds

Step-by-step explanation:

1 ton = 2000 pounds

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3 0
3 years ago
Help I’ll give 50 points<br> 12) 4x- y=8<br> 5x+y=1
Luba_88 [7]

12) 4x- y=8

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4 0
3 years ago
Factorise each of the following algebraic expressions completely,
LekaFEV [45]

Answer:

see explanation

Step-by-step explanation:

(a)

Given

2k - 6k² + 4k³ ← factor out 2k from each term

= 2k(1 - 3k + 2k²)

To factor the quadratic

Consider the factors of the product of the constant term ( 1) and the coefficient of the k² term (+ 2) which sum to give the coefficient of the k- term (- 3)

The factors are - 1 and - 2

Use these factors to split the k- term

1 - k - 2k + 2k² ( factor the first/second and third/fourth terms )

1(1 - k) - 2k(1 - k) ← factor out (1 - k) from each term

= (1 - k)(1 - 2k)

1 - 3k + 2k² = (1 - k)(1 - 2k) and

2k - 6k² + 4k³ = 2k(1 - k)(1 - 2k)

(b)

Given

2ax - 4ay + 3bx - 6by ( factor the first/second and third/fourth terms )

= 2a(x - 2y) + 3b(x - 2y) ← factor out (x - 2y) from each term

= (x - 2y)(2a + 3b)

8 0
3 years ago
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