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Murljashka [212]
2 years ago
15

The principal observes 8th grade the following day. Of the 120 students, 42 brought their lunch from home, 72 purchased lunches,

and 6 brought a lunch from home and purchased an ice cream.
Explain how the principal could use the numbers from 7th grade to predict the 8th grade possibilities.
Mathematics
1 answer:
Amanda [17]2 years ago
8 0

I think

Step-by-step explanation:

by finding each chart such as if there are 120 7th graders and 180 8th graders he can use the same numbers from the 7th graders and compare each percentage to the 8th-grade kids and get a close enough answer

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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
Please help! I can't figure this out
trasher [3.6K]

Answer:

see explanation

Step-by-step explanation:

(1)

CD = CE - DE = 8 - 2 = 6

CB = CD = 6 ( given as congruent ) , thus

AC = AB + CB = 4 + 6 = 10

(2)

∠ 2 + ∠ BCD + ∠ DCF  = 180° ( sum of angles on a straight line )

∠ 2 + 60° + 70° = 180°

∠ 2 + 130° = 180° ( subtract 130° from both sides )

∠ 2 = 50°

6 0
3 years ago
Read 2 more answers
Solve for p: 3/8 +p/8= 3 1/8
Ivenika [448]
Solve for p:
p/8 + 3/8 = 25/8
p/8 + 3/8 = (p + 3)/8:
(p + 3)/8 = 25/8

Multiply both sides of (p + 3)/8 = 25/8 by 8:
(8 (p + 3))/8 = (8×25)/8
(8×25)/8 = (8×25)/8:
(8 (p + 3))/8 = (8×25)/8
(8 (p + 3))/8 = 8/8×(p + 3) = p + 3:p + 3 = (8×25)/8
(8×25)/8 = 8/8×25 = 25:
p + 3 = 25

Subtract 3 from both sides:
p + (3 - 3) = 25 - 3
3 - 3 = 0:
p = 25 - 3
25 - 3 = 22:
Answer: p = 22
4 0
3 years ago
The opposite corners of a quadrilateral are right angles. The quadrilateral is not a rhombus. What kind of quadrilateral is this
tatiyna
A rhombus is a 4 sided shape with equal lengths. Even though it has equal lengths, it has opposite sides that are parallel, and angles that are equal. It also has diaginols (dashed lines) meat in the middle
8 0
3 years ago
-9(-5k+3m) +9m-5(-5m+9k)
liraira [26]

Answer:

the ansmwer is * grabs a notebook and doodle the equation* hmm.. it is 7m

happy to help!!!

3 0
3 years ago
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