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Levart [38]
3 years ago
8

What is the square root of -625

Mathematics
2 answers:
loris [4]3 years ago
6 0

Answer:

  25i

Step-by-step explanation:

Your calculator or your knowledge of powers of 5 will tell you that √625 is 25. The minus sign makes the root imaginary.

  \sqrt{-625}=\sqrt{(-1)(25^2)}=25\sqrt{-1}=25i

OLga [1]3 years ago
5 0

Answer:

6

Step-by-step explanation:

you welcome

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14. Solve: 3x² + 5x – 12 = 0<br>B. 4-3<br>C. 8, - 18​
VARVARA [1.3K]

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Step-by-step explanation:

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7 0
4 years ago
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3 years ago
Solve for X
nika2105 [10]

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masha68 [24]
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5 0
3 years ago
Let $f(x) = 2x^2 + 3x - 9,$ $g(x) = 5x + 11,$ and $h(x) = -3x^2 + 1.$ Find $f(x) - g(x) + h(x).$
Viefleur [7K]

QUESTION 1

Given that:

f(x)=2x^2+3x-9,

g(x)=5x+11,

and

h(x)=-3x^2+1

Then;

f(x)-g(x)+h(x)=2x^2+3x-9-(5x+11)+(-3x^2+1)

f(x)-g(x)+h(x)=2x^2+3x-9-5x-11-3x^2+1

Group similar terms;

f(x)-g(x)+h(x)=2x^2-3x^2+3x-5x-11-9+1

Simplify;

f(x)-g(x)+h(x)=-x^2-2x-19

QUESTION 2

Given that;

f(x)=4x-7.

g(x)=(x+1)^2

and

s(x)=f(x)+g(x)

Substitute the functions;

s(x)=4x-7+(x+1)^2

Substitute x=3

s(3)=4(3)-7+(3+1)^2

s(3)=12-7+(4)^2

s(3)=5+16

s(3)=21

QUESTION 3

Given:

f(x)=3x+2

g(x)=x^2-5x-1

f(g(x))=f(x^2-5x-1)

This implies that;

f(g(x))=3(x^2-5x-1)+2

Expand the parenthesis;

f(g(x))=3x^2-15x-3+2

f(g(x))=3x^2-15x-1

QUESTION 4

The given function is;

f(x)=3(x-6)^2+1

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y=3(x-6)^2+1

\Rightarrow y-1=3(x-6)^2

\Rightarrow \frac{y-1}{3}=(x-6)^2

\Rightarrow \sqrt{\frac{y-1}{3}}=x-6

\Rightarrow x=6+\sqrt{\frac{y-1}{3}}

The range is:

\frac{y-1}{3}\ge0

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y\ge1

The interval notation is;

[1,+\infty)

6 0
4 years ago
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