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Natalija [7]
3 years ago
6

Help me out please? Thank you

Mathematics
2 answers:
Luden [163]3 years ago
7 0

The correct answer is C.

Semenov [28]3 years ago
3 0

Parent function are the simplest function possible of a "function family".


See: for quadratic functions y = ax² + bx + c where a ≠ 0 all the others can be, so the simplest quadratic function would be y = ax² + 0x + c, or y = ax², but if we say a = 1 and just let y be equal to x²? We have the simplest of all y = x²


The other one who do this is y=\sqrt{x} he's the simplest function possible of root functions.

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RHOMBUS The diagonals of rhombus ABCD intersect at E. Given that
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Answer:

<u>Question 11:</u>

\angle DAC = 53^\circ

\angle AED = 90^\circ

\angle ADC = 74

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AE = 6.03

AC = 12.06

<u>Question 12:</u>

\triangle ABD, \triangle BAC, \triangle CDA and \triangle DAB

<u>Question 13: </u>

AC and BD are perpendicular lines, and they are diagonals

Step-by-step explanation:

<u>Question 11</u>

Given

\angle BAC = 53^\circ

DE = 8

See attachment for Rhombus

Required

Determine the indicated sides

Solving (a): \angle DAC

Diagonal CA divides \angle DAB into 2 equal angles

i.e

\angle DAC = \angle BAC

So:

\angle DAC = 53^\circ

Solving (b): \angle AED

The angles at E is 90 degrees because diagonals AC and BD meet at a perpendicular.

So:

\angle AED = 90^\circ

Solving (c): \angle ADC

First, we calculate \angle ADE, considering \triangle ADE:

\angle ADE + \angle AED + \angle DAC = 180

\angle ADE + 90 + 53 = 180

\angle ADE + 143 = 180

\angle ADE = -143 + 180

\angle ADE = 37

To calculate \angle ADC, we have:

\angle ADC = 2*\angle ADE

\angle ADC = 2* 37

\angle ADC = 74

Solving (d): DB

From the rhombus

DB = DE +EB

Where

DE =EB

So:

DB = 8 + 8

DB = 16

Solving (e): AE

To do this we consider \triangle ADE

Using the tan formula

tan(\angle ADE) = \frac{AE}{DE}

\angle ADE = 37 and DE = 8

So:

\tan(37) = \frac{AE}{8}

AE = 8 * \tan(37)

AE = 6.03

Solving (f): AC

This is calculated as:

AC = AE + EC

Where

AE = EC

AC = 6.03 +6.03

AC = 12.06

<u>Question 12: Isosceles Triangle</u>

In the rhombus, all 4 sides are equal;

So, the isosceles triangle are:

\triangle ABD, \triangle BAC, \triangle CDA and \triangle DAB

<u>Question 13: </u>

AC and BD are perpendicular lines, and they are diagonals

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