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3 years ago

What is the domain of the function y= (square root) x+4

Mathematics

1 answer:

garri49 [273]3 years ago

6 0

Answer:

0 ≤ x ≤ ∞

Step-by-step explanation:

Square roots always need to be either equal to or greater than zero because it should be positive inside (unless we're working with imaginary numbers).

So the domain would be that x ≥ 0 (or the third choice ⇒ 0 ≤ x ≤ ∞).

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ddd [48]

Answer:

the answer is b -2u +w

or -2w - u

8 0

3 years ago

(6.15x+0.24y)÷1.5+5.9x−1.6y Help plz

nadya68 [22]

Answer:

10x − 1.44y

Step-by-step explanation:

(6.15x + 0.24y) ÷ 1.5 + 5.9x − 1.6y =

= 4.1x + 0.16y + 5.9x − 1.6y

= 10x − 1.44y

8 0

3 years ago

Read 2 more answers

Help me solve these two questions please!

ozzi

They are simple linear equations with one unknown, lets tackle them, one step at the time solving for the unknown:
4x - 2(x - 5) = x + 13
4x - 2x + 10 = x + 13
2x + 10 = x + 13
2x - x = 13 - 10
x = 3
that is the solution.

3(6 - x) - 4 = 5x + 2(7x + 3)
18 - 3x - 4 = 5x + 14x + 6
14 - 3x = 19x + 6
14 - 6 = 19x + 3x
8 = 22x
x = 8/22 = 4/11

5 0

3 years ago

Please help please please please

Readme [11.4K]

It is a low posabilaty

8 0

3 years ago

If the parent function f(x) = x^2 is fatter translated 11 units to the left, then translated 5 units down, write the resulting f

nikklg [1K]

The quadratic function given by:

$f(x)=a(x-h)^2+k, \ \ \ a\neq 0$

is in vertex form. The graph of $f$ is a parabola whose axis is the vertical line $x=h$ and whose vertex is the point $(h, k)$ . So:

To translate the graph of a function to the right, left, upward or downward we have:

$For \ a \ positive \ real \ number \ c. \ \mathbf{Vertical \ and \ horizontal \ shifts} \\ in \ the \ graph \ of \ y=f(x) \ are \ represented \ as \ follows:\\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{upward}: \\ g(x)=f(x)+c \\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{downward}: \\ g(x)=f(x)-c \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ \mathbf{right}: \\ g(x)=f(x-c) \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ \mathbf{left}: \\ g(x)=f(x+c)$

By knowing this things, we can solve our problem as follows:

FIRST.

Translating 11 units to the left:

$g(x)=f(x+11) \\ \\ \therefore g(x)=(x+11)^2$

Then translating 5 units down:

$g(x)=f(x)-c \\ \\ \therefore g(x)=(x+11)^2-5$

Since the new function is fatter, the factor we need to multiply the term $(x+11)^2$ must be less than 1, to make the graph fatter. So, according to our options, there are two factors 1/2 and 2.

Therefore, the right answer is b. f(x) = 1/2(x + 11)^2 - 5

SECOND.

Translating 8 units to the right:

$g(x)=f(x-8) \\ \\ \therefore g(x)=(x-8)^2$

Then translating 1 unit down:

$g(x)=f(x)-c \\ \\ \therefore g(x)=(x-8)^2-1$

As explained in the previous case, there are two factors 1/3 and 3, so we choose the first one.

Therefore, the right answer is a. g(x) = 1/3(x - 8)^2 - 1

7 0

3 years ago