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3 years ago

11

A right rectangular prism is shown. What shape best describes the cross section cut perpendicular to the base of a right rectang

ular prism?

1 answer:

bulgar [2K]3 years ago

3 0

C a rectangle. brainily says this is too short so I am writing more

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Find the midpoint of endpoints -6 and -10

Fed [463]

The midpoint would be -8 because it is 2 away from both ends.

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2 years ago

At Ezra's restaurant and bakery, each cake is cut into 9 pieces. Let c represent the number of cakes and p represent the total p

zmey [24]

Answer:

C * 9 = P

9C = P

Step-by-step explanation:

Each cake is 9 pieces

Cakes times 9 equals pieces

c * 9 = P

3 0

3 years ago

Need help ASAP hshehebsbeb

Ivan

Answer:

I thing option A. all real numbers.

but IAM not exactly sure....

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3 years ago

Read 2 more answers

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

2 years ago

The question and answers are in this photo <br><br> Thank you for the help!

arlik [135]

Answer:

A

Step-by-step explanation

Just pick a

5 0

3 years ago

Read 2 more answers