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vampirchik [111]
2 years ago
7

Conner and Jana are multiplying (3⁵6⁸)(3⁹6¹⁰).

Mathematics
1 answer:
ad-work [718]2 years ago
7 0
<h3>Conner work is correct. Jana work is wrong</h3>

<em><u>Solution:</u></em>

<em><u>Given that,</u></em>

<em><u>Conner and Jana are multiplying:</u></em>

(3^56^8)(3^96^{10})

Given Conner's work is:

(3^56^8)(3^96^{10}) = 3^{5+9}6^{8+10} = 3^{14}6^{18}

We have to check if this work is correct

Yes, Conner work is correct

From given,

(3^56^8)(3^96^{10})\\\\3^5 \times 6^8 \times 3^9 \times 6^{10}

Use the following law of exponent

a^m \times a^n = a^{m+n}

Therefore,

3^5 \times 6^8 \times 3^9 \times 6^{10} = 3^5 \times 3^9 \times 6^8 \times 6^{10} = 3^{5+9} \times 6^{8+10} = 3^{14} \times 6^{18}

<em><u>Given Jana's work is:</u></em>

(3^56^8)(3^96^{10}) = 3^{5.9}6^{8.10} = 3^{45}6^{80}

This is incorrect

The powers of same base has to be added. But here, powers are multiplied which is wrong

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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°.

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