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PSYCHO15rus [73]
3 years ago
10

All of the following relations are functions, except A) y = x2. B) x = y2. C) y = |x|. D) y = x .

Mathematics
2 answers:
elena-s [515]3 years ago
7 0
B) x = y^2 because it fails the vertical line test making it not a function
yulyashka [42]3 years ago
4 0

The answer is <u>B</u> because.....

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Hello help me with this question thanks in advance​
Ede4ka [16]

\bold{\huge{\green{\underline{ Solutions }}}}

<h3><u>Answer </u><u>1</u><u>1</u><u> </u><u>:</u><u>-</u></h3>

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

\sf{HM = 5 cm }

  • <u>In </u><u>square </u><u>all </u><u>sides </u><u>of </u><u>squares </u><u>are </u><u>equal </u>

<u>The </u><u>perimeter </u><u>of </u><u>square </u>

\sf{ = 4 × side }

\sf{ = 4 × 5 }

\sf{ = 20 cm }

Thus, The perimeter of square is 20 cm

Hence, Option C is correct .

<h3><u>Answer </u><u>1</u><u>2</u><u> </u><u>:</u><u>-</u></h3>

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

\sf{MX  = 3.5 cm }

  • <u>In </u><u>square</u><u>,</u><u> </u><u>diagonals </u><u>are </u><u>equal </u><u>and </u><u>bisect </u><u>each </u><u>other </u><u>at </u><u>9</u><u>0</u><u>°</u>

<u>Here</u><u>, </u>

\sf{MX  = MT/2}

\sf{MT = 2 * 3.5 }

\sf{MT = 7 cm}

Thus, The MT is 7cm long

Hence, Option C is correct .

<h3><u>Answer </u><u>1</u><u>3</u><u> </u><u>:</u><u>-</u><u> </u></h3>

<u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>measure </u><u>of </u><u>Ang</u><u>l</u><u>e</u><u> </u><u>MAT</u>

  • <u>All </u><u>angles </u><u>of </u><u>square </u><u>are </u><u>9</u><u>0</u><u>°</u><u> </u><u>each </u>

<u>From </u><u>above </u>

\sf{\angle{MAT  = 90° }}

Thus, Angle MAT is 90°

Hence, Option B is correct .

<h3><u>Answer </u><u>1</u><u>4</u><u> </u><u>:</u><u>-</u></h3>

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

  • <u>All </u><u>the </u><u>angles </u><u>of </u><u>square </u><u>are </u><u>equal </u><u>and </u><u>9</u><u>0</u><u>°</u><u> </u><u>each </u>

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

\sf{\angle{MHA  = }}{\sf{\angle{ MHT/2}}}

\sf{\angle{MHA = 90°/2}}

\sf{\angle {MHA = 45°}}

Thus, Angle MHA is 45°

Hence, Option A is correct

<h3><u>Answer </u><u>1</u><u>5</u><u> </u><u>:</u><u>-</u><u> </u></h3>

Refer the above attachment for solution

Hence, Option A is correct

<h3><u>Answer </u><u>1</u><u>6</u><u> </u><u>:</u><u>-</u><u> </u></h3>

Both a and b

  • <u>The </u><u>median </u><u>of </u><u>isosceles </u><u>trapezoid </u><u>is </u><u>parallel </u><u>to </u><u>the </u><u>base</u>
  • <u>The </u><u>diagonals </u><u>are </u><u>congruent </u>

Hence, Option C is correct

<h3><u>Answer </u><u>1</u><u>7</u><u> </u><u>:</u><u>-</u></h3>

In rhombus PALM,

  • <u>All </u><u>sides </u><u>and </u><u>opposite </u><u>angles </u><u>are </u><u>equal </u>

Let O be the midpoint of Rhombus PALM

<u>In </u><u>Δ</u><u>OLM</u><u>, </u><u>By </u><u>using </u><u>Angle </u><u>sum </u><u>property </u><u>:</u><u>-</u>

\sf{35° + 90° + }{\sf{\angle{ OLM = 180°}}}

\sf{\angle{OLM = 180° - 125°}}

\sf{\angle{ OLM = 55° }}

<u>Now</u><u>, </u>

\sf{\angle{OLM = }}{\sf{\angle{OLA}}}

  • <u>OL </u><u>is </u><u>the </u><u>bisector </u><u>of </u><u>diagonal </u><u>AM</u>

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

\sf{\angle{ PLA = 55° }}

Thus, Angle PLA is 55° .

Hence, Option C is correct

8 0
3 years ago
Find the value of x<br>​
Monica [59]

X= 130°

Step-by-step explanation:

From E, draw EF || AB || CD.

Now, EF || CD and CE is the transversal.

So; <DCE + <CEF = 180° [co. int. <s]

x° + <CEF = 180°

<CEF = (180° – x°).

Again, EF || AB and AE is the transversal.

So; <BAE + <AEF = 180° 01 [co. int. <s ]

105° + <AEC+ <CEF = 180°

105° +25° + (180° – x°) = 180°

x° = 130°

I hope I helped you^_^

7 0
3 years ago
Out of 200 people eating at a diner, 70% ordered
scZoUnD [109]

Answer:

140 people

Step-by-step explanation:

4 0
2 years ago
MATH<br> MATh<br> MAtH<br> MAAAAThHhH<br> YAAA
MariettaO [177]

Answer:

6k + 19

Step-by-step explanation:

7k - k + 19

6k + 19

7 0
2 years ago
Read 2 more answers
NEED HELP ASAP ‼️‼️I BELIEVE ANSWER IS C
shutvik [7]

Answer:

I agree with your answer C!!

Step-by-step explanation:

5 0
3 years ago
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