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Alenkasestr [34]

4 years ago

6

Mrs . Haider went to the mall and purchased tacos and burritos at the food court. 20 of her family member were visiting, so she

had to buy lots of food. tacos were $2.00 and burritos were $3.50. if Ms. hider needs to buy three times as many tacos as burritos and wants to spend $ 134.56.
A.) if b represent the number of burritos that she bought, write an equation to represent this scenario using only one variable.
b.) how many burritos can she buy?

1 answer:

-BARSIC- [3]4 years ago

7 0

the correct answer to your problem is a

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PtichkaEL [24]

2800 is the mass of the truck

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3 years ago

Please show the working and answer. you can take a picture for the working.

baherus [9]

Answer:

(a) The area of the triangle is approximately 39.0223 cm²

(b) ∠SQR is approximately 55.582°

Step-by-step explanation:

(a) By sin rule, we have;

SQ/(sin(∠SPQ)) = PQ/(sin(∠PSQ)), which gives;

5.4/(sin(52°)) = 6.8/(sin(∠PSQ))

∴ (sin(∠PSQ)) = (6.8/5.4) × (sin(52°)) ≈ 0.9923

∠PSQ = sin⁻¹(0.9923) ≈ 82.88976°

Similarly, we have;

5.4/(sin(52°)) = SP/(sin(180 - 52 - 82.88976))

Where, 180 - 52 - 82.88976 = ∠PQS = 45.11024

SP = 5.4/(sin(52°))×(sin(180 - 52 - 82.88976)) ≈ 4.8549

Given that RS : SP = 2 : 1, we have;

RS = 2 × SP = 2 × 4.8549 ≈ 9.7098

We have by cosine rule, $\overline {RQ}$ ² = $\overline {SQ}$ ² + $\overline {SR}$ ² - 2 × $\overline {SQ}$ × $\overline {SR}$ × cos(∠QSR)

∠QSR and ∠PSQ are supplementary angles, therefore;

∠QSR = 180° - ∠PSQ = 180° - 82.88976° = 97.11024°

∠QSR = 97.11024°

Therefore;

$\overline {RQ}$ ² = 5.4² + 9.7098² - 2 × 5.4×9.7098× cos(97.11024)

$\overline {RQ}$ ² ≈ 136.42

$\overline {RQ}$ = √(136.42) ≈ 11.6799

The area of the triangle = 1/2 × $\overline {PQ}$ × $\overline {PR}$ × sin(∠SPQ)

By substituting the values, we have;

1/2 × $\overline {PQ}$ × $\overline {PR}$ × sin(∠SPQ)

1/2 × 6.8 × (4.8549 + 9.7098) × sin(52°) ≈ 39.0223 cm²

The area of the triangle ≈ 39.0223 cm²

(b) By sin rule, we have;

$\overline {RS}$ /(sin(∠SQR)) = $\overline {RQ}$ /(sin(∠QSR))

By substituting, we have;

9.7098/(sin(∠SQR)) = 11.6799/(sin(97.11024))

sin(∠SQR) = 9.7098/(11.6799/(sin(97.11024))) ≈ 0.82493

∠SQR = sin⁻¹(0.82493) ≈ 55.582°.

8 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

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