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aalyn [17]
3 years ago
10

HELP!!! What’s the domain and range? PLS!!!!

Mathematics
1 answer:
Arte-miy333 [17]3 years ago
7 0

Answer:

A) Vertex: (3,1) Maximum, Domain: all real numbers, Range: y is less than or equal to 1.

B) Vertex: (1,-5) Minimum, Domain: all real numbers, Range: y is greater than or equal to -5

C) Vertex: (0,1) Minimum, Domain: all real numbers, Range: y is greater than or equal to 1

D) Vertex: (4,0) Maximum, Domain: all real numbers, Range: y is less than or equal to 0.

E) Vertex: (4,1) Minimum, Domain: all real numbers, Range: y is greater than or equal to 1

F) Vertex: (-1,4) Maximum, Domain: all real numbers, Range: y is less than or equal to 4.

G) Vertex: (2,-4) Minimum, Domain: all real numbers, Range: y is greater than or equal to -4

H) Vertex: (-4,-3) Minimum, Domain: all real numbers, Range: y is greater than or equal to -3

I) Vertex: (2,3) Maximum, Domain: all real numbers, Range: y is less than or equal to 3.

Step-by-step explanation:

Sorry it took forever to type

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Solve for x given that QRST is a rhombus.​
AVprozaik [17]

\bold{\huge{\underline{ Solution }}}

<h3><u>Given </u><u>:</u><u>-</u></h3>

  • We have given Rhombus QRST
  • The measurement of two sides of rhombus that is ( 6x - 5 ) and ( 4x + 13).

<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>value </u><u>of </u><u>x </u>

<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u></h3>

<u>Here</u><u>, </u><u> </u><u>We </u><u>have </u>

  • Rhombus QRST

<u>We </u><u>know </u><u>that</u><u>, </u>

  • All sides of rhombus are equal
  • Opposite sides of rhombus are parallel to each other

<u>Therefore</u><u>, </u>

<u>From </u><u>the </u><u>figure</u><u>, </u><u>we </u><u>can </u><u>conclude </u><u>that </u><u>:</u><u>-</u>

\bold{ 6x - 5 = 4x + 13 }

\sf{ 6x  = 4x + 13 + 5 }

\sf{ 6x -  4x =  13 + 5 }

\sf{ 2x =  18 }

\sf{ x = }{\sf{\dfrac{18}{2}}}

\bold{ x = 9 }

Hence, The value of x is 9 .

<h3><u>The </u><u>measurement </u><u>of </u><u>the </u><u>sides </u><u>of </u><u>rhombus </u><u>QRST</u><u> </u><u>:</u><u>-</u><u> </u></h3>

Length of side QR

\sf{ = 6x - 5 }

\sf{ = 6(9) - 5 }

\sf{ = 54 - 5 }

\bold{ = 49 }

Length of side RS

\sf{ = 4x + 13  }

\sf{ = 4(9) + 13 }

\sf{ = 36 + 13  }

\bold{ = 49 }

<u>From </u><u>above </u><u>we </u><u>can </u><u>say </u><u>that</u><u>, </u>

  • All sides of rhombus are equal.

<h3><u>Some </u><u>more </u><u>properties </u><u>of </u><u>rhombus </u></h3>

  • Opposite angles of rhombus are equal
  • Diagonals of rhombus bisect each other at 90°
  • Diagonals of rhombus also bisect the side angles of rhombus
  • Rhombus is a quadrilateral, so the sum of interior angles of rhombus are 360°
  • Whereas, The sum of the adjacent angles of rhombus are 180°.
  • All rhombus are parallelogram but all parallelograms are not rhombus .
5 0
2 years ago
What’s the correct answer for this question?
jarptica [38.1K]

Answer:

6

Step-by-step explanation:

In the given figure, secants PA and PC are intersecting outside of the circle at point P.

PA = PB + AB = 7 + 5 = 12

PC = 14, PD =?

By the property of intersecting secants outside of a circle:

PB \times PA = PD\times PC\\\therefore 7\times 12 = PD \times 14\\\therefore 84 =PD \times 14\\\\\therefore PD = \frac{84}{14}\\\\\therefore PD = 6\\\\

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Divide. Enter the quotient in simplest form.
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Its should be 0 if there negative or 12-
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