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Marta_Voda [28]
2 years ago
5

X² = -25 Solve the quadratic equation using the square root method

Mathematics
2 answers:
mario62 [17]2 years ago
5 0

Answer:

x= maths error

Step-by-step explanation:

x² = -25

x=√-25

x= maths error

Karo-lina-s [1.5K]2 years ago
5 0
Answer = No Real Solutions
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UkoKoshka [18]

Answer:

-14°F

Step-by-step explanation:

First, we start at 12°f. Then, the temperature changes by -26°f. This means that we add (-26°f) to what we currently have. We then have

12 + (-26)

= 12-26

= -14°F

5 0
3 years ago
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If the perimeters of each shape are equal, which equation can be used to find the value of x? A triangle with base x + 2, height
Tju [1.3M]

Correct question :

If the perimeters of each shape are equal, which equation can be used to find the value of x? A triangle with base x + 2, height x, and side length x + 4. A rectangle with length of x + 3 and width of one-half x. (x + 4) + x + (x + 2) = one-half x + (x + 3) (x + 2) + x + (x + 4) = 2 (one-half x) + 2 (x + 3) 2 (x) + 2 (x + 2) = 2 (one-half x) + 2 (x + 3) x + (x + 2) + (x + 4) = 2 (x + 3 and one-half)

Answer: (x + 2) + x + (x + 4) = 2 (one-half x) + 2 (x + 3)

Step-by-step explanation:

Given the following :

A triangle with base x + 2, height x, and side length x + 4 - - - -

b = x + 2 ; a = x ; c = x + 4

Perimeter (P) of a triangle :

P = a + b + c

P =( x + 2) + x + (x + 4) - - - (1)

A rectangle with length of x + 3 and width of one-half x

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Perimeter of a rectangle (P) = 2(l+w)

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If perimeter of each same are the same ; then;

(1) = (2)

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7 0
3 years ago
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Paul [167]

Answer:

1

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9

16

25

36

49

64

81

100

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9 = 3 x 3 = 3²

16 = 4 x 4 = 4²

25 = 5 x 5 = 5²

36 = 6 x 6 = 6²

49 = 7 x 7 = 7²

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4 0
2 years ago
PLEASE HELP!
hammer [34]

Answer:

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Step-by-step explanation:

Taken together, the triangles form a square whose side length is 4. The area is ...

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6 0
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Please help me out!!!!!
Gnoma [55]
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When you graph this function, it crosses the x-axis once. In other words, that means there is only 1 real zero and 2 imaginary complex zeros. In addition, imaginary solutions always come in pairs, so there can’t be a odd number of them such as 3.
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