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3 years ago

11

Square Root Functions Help!

2 answers:

Montano1993 [528]3 years ago

5 0

Answer:

$f(x)=\sqrt{x-1}$

Step-by-step explanation:

Hello there!

I take it you are asking for the expression of the function shown.

Since it is a normal square root function moved 1 unit to the right, you will subtract 1 from x.

:)

Papessa [141]3 years ago

5 0

Answer:

y = √(x - 1)

Step-by-step explanation:

We start with the basic square root function, whose domain is [0, ∞ ]. The given graph is that of this basic square root function, EXCEPT that the whole graph has been translated 1 unit to the right.

The graph shown is that of y = √(x - 1).

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Someone please help me in math

Helga [31]

I just solved it and I'm pretty sure its only one solution, I took the bottom equation multiplied it by -3 then i canceled out the equations and got 9. So that would make it C.

3 0

3 years ago

Read 2 more answers

Please please help me no links

Finger [1]

Given:

In triangle ABC, AB = AC, AD is angle bisector and measure of angle C is 49 degrees.

To find:

The value of x and y.

Solution:

In triangle ABC,

$AB=AC$ (Given)

So, triangle ABC is an isosceles triangle and by the definition of base angles the base angles of isosceles triangle are congruent.

In isosceles triangle ABC,

$\angle B\cong \angle C$

$m\angle B\cong m\angle C$

$m\angle B\cong 49^\circ$

The angle bisector of an isosceles triangle is the median and altitude of the triangle. So, the angle bisector is perpendicular to the base.

$m\angle ADB=90^\circ$

$x^\circ=90^\circ$

In triangle ABD,

$m\angle DAB+m\angle ABD+m\angle ADB=180^\circ$ [Angle sum property]

$y^\circ+49^\circ+x^\circ=180^\circ$

$y^\circ+49^\circ+90^\circ=180^\circ$

$y^\circ=180^\circ-49^\circ-90^\circ$

$y^\circ=41^\circ$

Therefore, the correct option is B.

3 0

3 years ago

This hyperbola is centered at theorigin. Find its equation.Foci: (-2,0) and (2,0)Vertices: (-1,0) and (1,0)

beks73 [17]

SOLUTION

From the question, the center of the hyperbola is

$\begin{gathered} (h,k),\text{ which is } \\ (0,0) \end{gathered}$

a is the distance between the center to vertex, which is -1 or 1, and

c is the distance between the center to foci, which is -2 or 2.

b is given as

$\begin{gathered} b^2=c^2-a^2 \\ b^2=2^2-1^2 \\ b=\sqrt[]{3} \end{gathered}$

But equation of a hyperbola is given as

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$

Substituting the values of a, b, h and k, we have

$\begin{gathered} \frac{(x-0)^2}{1^2}-\frac{(y-0)^2}{\sqrt[]{3}^2}=1 \\ \frac{x^2}{1}-\frac{y^2}{3}=1 \end{gathered}$

Hence the answer is

$\frac{x^2}{1}-\frac{y^2}{3}=1$

5 0

8 months ago

If you spend $60 for dinner at the restaurant, what should be your 15% tip?

LUCKY_DIMON [66]

Answer:

.15 * 60 = 9

$9

Step-by-step explanation:

5 0

2 years ago

frez [133]

I GOT YOU

1. b

2. b

3. b

4. c

5. c

6. a

7. c

8. b

9. a

10. d

hope this helps!

6 0

3 years ago

Read 2 more answers