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Lady bird [3.3K]
3 years ago
7

What value is a discontinuity of x squared plus 5 x plus 2, all over x squared plus 2 x minus 35?

Mathematics
2 answers:
iVinArrow [24]3 years ago
6 0
The value of discontinuity is x= -7, x= 5
Aleksandr-060686 [28]3 years ago
6 0

Answer:

x=-7 and x=5.  

Step-by-step explanation:

We have been given a rational function: f(x)=\frac{x^2+5x+2}{x^2+2x-35}. We are asked to find the points at which our function is discontinuous.

A rational function is discontinuous when the function is undefined or the denominator is zero.

Let us find what values of x will make our denominator zero.

x^2+2x-35=0

We will use factoring to find the zeros of x. By splitting the middle term we will get,

x^2+7x-5x-35=0

x(x+7)-5(x+7)=0

(x+7)(x-5)=0

(x+7)=0\text{ or }(x-5)=0

x=-7\text{ or }x=5

Therefore, at x equals -7 and x equals 5 our function is discontinuous.

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Step-by-step explanation:

If we look at system 1, we see that if we multiply the first equation by 2 and subtract the y value so that its on the same side as x, the equations are the same. That means that no matter the x or y value, it will be a solution.

If we look at system 2, however, we see that if we multiply the top equation by 3, the equations are the exact same except for the fact that the top equation has a -21. This means that no matter the x or y, there will never be a solution.

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3 years ago
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

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