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Kazeer [188]
2 years ago
11

Which inequality describes the acceptable weights for the suitcase?

Mathematics
1 answer:
Yanka [14]2 years ago
6 0

Based on the given options, the inequality for the acceptable weights of the suitcase is 32.5 ≤ x ≤ 47.5.

<h3>What is the acceptable weight for the suitcase?
</h3>

You didn't include the full question but this question can be answered based on the given options.

When inequalities are written, there is no inclusion of "or" which eliminates options A and C.

It is illogical to have an inequality as 32.5 ≥ x ≥ 47.5. This is because if 32.5 is larger or equal to x, then x cannot be larger than 47.5. This is mathematically impossible.

This leaves the likely answer as 32.5 ≤ x ≤ 47.5 which means that the weight of the suitcase has to be larger than or equal to 32.5 but less than or equal to 47.5.

Find out more on inequalities at brainly.com/question/24372553.

#SPJ1

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3 years ago
What’s the 5th term is 2,14,98
iragen [17]

\bold{\huge{\orange{\underline{ Solution }}}}

<h3><u>Correct </u><u>Question </u><u>:</u><u>-</u></h3>

What is the 5th term of an AP 2 , 14 ....98 .

<h3><u>Given </u><u>:</u><u>-</u><u> </u></h3>

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

  • \sf{ 2 , 14 ,.... 98 }
  • <u>AP </u><u>is </u><u>the </u><u>arithmetic </u><u>progression </u><u>or </u><u>a </u><u>sequence </u><u>of </u><u>numbers </u><u>in </u><u>which </u><u>succeeding </u><u>number </u><u>is </u><u>differ </u><u>from </u><u>preceeding </u><u>number </u><u>by </u><u>a </u><u>common </u><u>value</u><u>. </u>

<h3><u>Solution </u><u>:</u><u>-</u></h3>

<u>We </u><u>have </u><u>an </u><u>AP </u><u>:</u><u>-</u><u> </u><u>2</u><u> </u><u>,</u><u> </u><u>1</u><u>4</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>9</u><u>8</u>

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

  • \sf{a1\: or \:1st\: term = 2 }
  • \sf{a2 \:or \:2nd\: term = 14 }
  • \sf{an \:or\: last\: term = 98}

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

Common difference of an AP

\sf{ a2 - a1 }

\sf{ = 14 - 2}

\sf{ = 12}

Thus, The common difference is 12

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

We know that,

\sf{an = a1 + ( n - 1 )d}

\sf{a5 = 2 + ( 5 - 1 )12}

\sf{a5 = 2 + 4 × 12}

\sf{a5 = 2 + 48}

\sf{\red{a5 = 50}}

Hence, The 5th term of given AP is 50

4 0
2 years ago
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