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2 years ago

Use the table to write a proportion

Mathematics

1 answer:

Marizza181 [45]2 years ago

7 0

Answer:

n = 17

$\frac{n}{85}= \frac{34}{170}$

Step-by-step explanation:

$\frac{n}{85}= \frac{34}{170}$

n · 170 = 34 · 85

170y = 2890

170y ÷ 170 = 2890 ÷ 170

y = 17

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Peter says if you subtract 13 from my number and multiply the difference by -6, the result is -42. What is Peters's number?

qwelly [4]

I think the answer is 20, may be wrong tho

3 0

2 years ago

Read 2 more answers

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

2 years ago

1. Two object accumulated a charge of

In-s [12.5K]

Answer:

The magnitude of the electric force between these two objects

will be: 181.274 N.

i.e. F $=181.274$ N

Step-by-step explanation:

Two object accumulated a charge of 4.5 μC and another a charge of 2.8 μC.

q₁ = 4.5 μC = 4.5 × 10⁻⁶ C

q₂ = 2.8 μC = 2.8 × 10⁻⁶ C

separated distance = d = 2.5 cm

Calculating the magnitude of the force between two charged objects using the formula:

$F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}$

$=\:\frac{4.5\times \:\:10^{-6}\:\times \:\:2.8\:\times \:\:10^{-6}}{4\:\times \:\left(3.14\right)\:\times \:\left(8.85\times \:10^{-12}\right)\times \left(2.5\times \:\:10^{-2}\right)^2}$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:4\times \:8.85\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $4.5\times \:10^{-6}\times \:2.8\times \:10^{-6}=10^{-12}\times \:12.6$

$=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:35.4\pi \left(10^{-2}\times \:2.5\right)^2}$ ∵ $\mathrm{Multiply\:the\:numbers:}\:4\times \:8.85=35.4$