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Anna11 [10]
2 years ago
11

11. What is the domain of the function? Please show how you solved it and explain.

Mathematics
1 answer:
joja [24]2 years ago
8 0

Answer:

domain:  x\leq -7, x\geq 3

Step-by-step explanation:

f(x)=\sqrt{x^2+4x-21}

Factor x^2+4x-21:

x^2+4x-21=(x-3)(x+7)

Therefore,

f(x)=\sqrt{(x-3)(x+7)}

\implies f(x)=\sqrt{(x-3)}\sqrt{(x+7)}

Find non-negative values for radicals:

\sqrt{(x-3)} \implies x\geq 3

\sqrt{(x+7)} \implies x\leq -7

Therefore, domain:  x\leq -7, x\geq 3

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Kryger [21]

Answer:

1)

Given the triangle RST with Coordinates  R(2,1), S(2, -2) and T(-1 , -2).

A dilation is a transformation which produces an image that is the same shape as original one, but is different size.  

Since, the scale factor \frac{5}{3} is greater than 1, the image is enlargement or a stretch.  

Now, draw the dilation image of the triangle RST with center (2,-2) and scale factor \frac{5}{3}

Since, the center of dilation at S(2,-2) is not at the origin, so the point S and its image S{}' are same.

Now, the distances from the center of the dilation at point S to the other points R and T.  

The dilation image will be\frac{5}{3} of each of these distances,

SR=3, so S{}'R{}'=5 ;


ST=3, so S{}'T{}'=5  

Now, draw the image of RST i.e R'S'T'

Since, RT=3\sqrt{2} [By using hypotenuse of right angle triangle] and R{}'T{}'=5\sqrt{2}.


2)

(a)

Disagree with the given statement.

Side Angle Side postulate (SAS) states that:

If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle then these two triangles are congruent.

Given: B is the midpoint of \overline{AC} i.e \overline{AB}\cong \overline{BC}

In the triangle ABD and triangle CBD, we have

\overline{AB}\cong \overline{BC}   (SIDE)            [Given]

\overline{BD}\cong \overline{BD}   (SIDE)            [Reflexive post]

Since, there is no included angle in these triangles.

∴ \Delta ABD is not congruent to \Delta CBD .

Therefore, these triangles does not follow the SAS congruence postulates.

(b)

SSS(SIDE-SIDE-SIDE) states that if three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent.

Since it is also given that  \overline{AD}\cong \overline{CD}.

therefore, in the triangle ABD and triangle CBD, we have

\overline{AB}\cong \overline{BC}   (SIDE)            [Given]

\overline{AD}\cong \overline{CD}   (SIDE)           [Given]

\overline{BD}\cong \overline{BD}   (SIDE)            [Reflexive post]

therefore by, SSS postulates \Delta ABD\cong \Delta CBD.

3)

Given that:  \angle1=\angle 3 are vertical angles, as they are formed by intersecting lines.

Therefore

, by the definition of linear pairs

\angle 1 and \angle 2 and \angle 3  and \angle 2 are linear pair.

By linear pair theorem, \angle 1 and \angle 2   are supplementary, \angle 2 and \angle 3  are supplementary.

m\angle1+m\angle 2=180^{\circ}

m\angle2+m\angle 3=180^{\circ}

Equate the above expressions:

m\angle 1+m\angle 2=m\angle 2+m\angle 3

Subtract the angle 2 from both sides in the above expressions

∴m\angle 1=m\angle 3

By Congruent Supplement theorem: If two angles are supplements of the same angle, then the two angles are congruent.


therefore, \angle 1\cong \angle 3.















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