What is the domain and range of this function?

ANSWER FAST PLEASE

NemiM [27]

Answer:

B.18

Step-by-step explanation:

2+1+3

3+3=6

6+5=11

11+7=18

6 0

3 years ago

Read 2 more answers

What is the domain of the function f(x) = x^2 + 5?

Sophie [7]

The domain is R so the first one is correct.

4 0

3 years ago

The grid below shows figure Q and its image figure Q' after a transformation:

Masteriza [31]

<h3>Answer: choice B) counterclockwise rotation of 90 degrees around the origin</h3>

To go from figure Q to figure Q', we rotate one of two ways

* 270 degrees clockwise

* 90 degrees counterclockwise

Since "270 clockwise" isn't listed, this means "90 counterclockwise" is the only possibility.

4 0

3 years ago

Given: △ABC, AB=5sqrt2 m∠A=45°, m∠C=30° Find: BC and AC

Marysya12 [62]

BC is 10 units and AC is $5+5\sqrt{3}$ units

Step-by-step explanation:

Let us revise the sine rule

In ΔABC:

$\frac{AB}{sin(C)}=\frac{BC}{sin(A)}=\frac{AC}{sin(B)}$
AB is opposite to ∠C
BC is opposite to ∠A
AC is opposite to ∠B

Let us use this rule to solve the problem

In ΔABC:

∵ m∠A = 45°

∵ m∠C = 30°

- The sum of measures of the interior angles of a triangle is 180°

∵ m∠A + m∠B + m∠C = 180

∴ 45 + m∠B + 30 = 180

- Add the like terms

∴ m∠B + 75 = 180

- Subtract 75 from both sides

∴ m∠B = 105°

∵ $\frac{AB}{sin(C)}=\frac{BC}{sin(A)}$

∵ AB = $5\sqrt{2}$

- Substitute AB and the 3 angles in the rule above

∴ $\frac{5\sqrt{2}}{sin(30)}=\frac{BC}{sin(45)}$

- By using cross multiplication

∴ (BC) × sin(30) = $5\sqrt{2}$ × sin(45)

∵ sin(30) = 0.5 and sin(45) = $\frac{1}{\sqrt{2}}$

∴ 0.5 (BC) = 5

- Divide both sides by 0.5

∴ BC = 10 units

∵ $\frac{AB}{sin(C)}=\frac{AC}{sin(B)}$

- Substitute AB and the 3 angles in the rule above

∴ $\frac{5\sqrt{2}}{sin(30)}=\frac{AC}{sin(105)}$

- By using cross multiplication

∴ (AC) × sin(30) = $5\sqrt{2}$ × sin(105)

∵ sin(105) = $\frac{\sqrt{6}+\sqrt{2}}{4}$

∴ 0.5 (AC) = $\frac{5+5\sqrt{3}}{2}$

- Divide both sides by 0.5

∴ AC = $5+5\sqrt{3}$ units

BC is 10 units and AC is $5+5\sqrt{3}$ units

Learn more:

You can learn more about the sine rule in brainly.com/question/12985572

#LearnwithBrainly

6 0

3 years ago

Polygon CCC has an area of 404040 square units. K 2ennan drew a scaled version of Polygon CCC using a scale factor of \dfrac12 1

nikdorinn [45]

Answer:

Area of polygon D = 10 square units

Step-by-step explanation:

Given:

Polygon C has an area of 40 square units.

It is scaled with a scale factor of $\frac{1}2$ to form a new polygon D.

To find:

The area of polygon D = ?

Solution:

When any polygon is scaled to half, then all the sides of new polygon are half of the original polygon.

And the area becomes one-fourth of the original polygon.

Let us consider this by taking examples:

First of all, let us consider a right angled triangle with sides 6, 8 and 10 units.

Area of a right angled triangle is given by:

$A = \dfrac{1}{2} \times Base \times Height\\\Rightarrow A = \dfrac{1}{2} \times 6 \times 8 = 24\ sq\ units$

If scaled with a factor $\frac{1}{2}$ , the sides will be 3, 4 and 5.

New area, A':

$A' =\dfrac{1}{2} \times 3 \times 4 = 6\ sq\ units = \dfrac{1}4\times A$

i.e. Area becomes one fourth.

Let us consider a rectangle now.

Sides be 8 and 10 units.

Area of a rectangle, A = $Length \times Width$ = 8 $\times$ 10 = 80 sq units.

Now after scaling, the sides will be 4 and 5 units.

New Area, A' = 4 $\times$ 5 =20 sq units

So, $\bold{A' = \frac{1}4 \times A}$

Now, we can apply the same in the given question.

$\therefore$ Area of polygon D = $\bold{\frac{1}{4}}$ $\times$ Area of polygon C

Area of polygon D = $\bold{\frac{1}{4}}$ $\times$ 40 = 10 sq units

4 0

3 years ago

What is the domain and range of this function?​

What is the domain and range of this function?