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Murrr4er [49]
3 years ago
13

Help! My friends needs help rlly bad

Mathematics
1 answer:
Nitella [24]3 years ago
4 0

The answer to number 1 is 2/9 of her paycheck. The answer to number 2 is 3/5. Its square is 9/25 which is in between 1/3 and 1/2. It's reciprocal is 5/3 which is in between 1 and 2. It is also in simplest form.

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PLEASE HURRY!!!!!!!!!!!!
den301095 [7]

Answer:

b cause 3.0 to 4.5 squared to mc4- 9280=gawk gawk 9000

5 0
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

8 0
3 years ago
Can I have help solving this?
tensa zangetsu [6.8K]

4^2 + 2(4 + 6)

4^2 + 2(10)

16 + 2(10)

16 + 20

36

3 0
4 years ago
Please please please help me out
Murrr4er [49]

Answer:

what grade you in do you even do your work this easy is b

8 0
3 years ago
Read 2 more answers
Determine whether the triangles are congruent. Explain your reasoning .
Vanyuwa [196]

Answer:

The triangles are congruent by congruency theorem SAS.

Step-by-step explanation:

Side-Angle-Side, as it suggests, is when a triangle is congruent by the order of congruent side, congruent angle, and another congruent side. If named certain ways, they can be congruent. Since the question does not provide a specific way of naming the triangles, we can assume that any way is allowed. 2.5 and 2.5 are congruent, followed by angles D and G which have the congruent angle markers. The congruent angles are followed by 1.7 and 1.7, making the triangles congruent by Side-Angle-Side.

7 0
3 years ago
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