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Luda [366]
3 years ago
14

Find the zeros for (x+7) (x+5)+0

Mathematics
1 answer:
JulijaS [17]3 years ago
3 0

Steps to solve for the zero:

(x + 7)(x + 5) = 0

~Set both factors to equal zero

x + 7 = 0

x + 5 = 0

~Solve for x in [ x + 7 = 0 ]

x + 7 = 0

x + 7 - 7 = 0 - 7

x = -7

~Solve for x in [ x + 5 = 0 ]

x + 5 = 0

x + 5 - 5 = 0 - 5

x = -5

Therefore, the solutions are x = -7 and x = -5

Best of Luck!

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Answer:

2.5 teaspoon.

Step-by-step explanation:

4 serve recipe need cinnamon = 1 tsp

1 serve recipe need cinnamon = 1/4 tsp

10 serve recipe need cinnamon = 1/4 * 10

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What types of numbers are undefined when they are under a radical sign? If you were dealing with the number √-1, would it be def
Misha Larkins [42]
The square root of a a negative integer is imaginary.
It would still be a negative under a square root if you multiplied it by 2, therefor it will still be imaginary, or I’m assuming as your book calls it, undefined.
2•(sqrt-1) = 2sqrt-1

If you add a number to -1 itself, specifically 1 or greater it will become a positive number or 0 assuming you just add 1. In that case it would be defined.
-1 + 1 = 0
-1 + 2 = 1

If you add a number to the entire thing “sqrt-1” it will not be defined.
(sqrt-1) + 1 = 1+ (sqrt-1)

If you subtract a number it will still have a negative under a square root, meaning it would be undefined.
(sqrt-1) + 1 = 1 + (sqrt-1)

however if you subtract a negative number from -1 itself, you end up getting a positive number or zero. (Subtracting a negative number is adding because the negative signs cancel out).
-1 - -1 = 0
-1 - -2 = 1

If you squared it you would get -1, which is defined
sqrt-1 • sqrt-1 = -1

and if you cubed it, you would get a negative under a square root again, therefor it would be undefined.
sqrt-1 • sqrt-1 • sqrt-1 = -1 • sqrt-1 = -1(sqrt-1)

Sorry if this answer is confusing, I don’t have a scientific keyboard, I’ll get one soon.
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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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