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kakasveta [241]
2 years ago
10

WILL GIVE BRAINLIEST

Mathematics
1 answer:
Lilit [14]2 years ago
8 0

Substitution involves plugging the given values of the variables and the

parenthesis is a sign for multiplication (multiplied by).

<h3>The correct responses are as follows;</h3><h3 />
  • 7. 2·(x + 4) + 10 = <u>28</u>

  • 8. 3·(4·x) + 4 =<u> 64</u>

  • 9. 10·(x·y) = <u>300</u>

  • 10. 2·(x + y) = <u>22</u>

  • 11. 3·r × s + 4·x = <u>38</u>
  • 12. 2×q + 2×x + 2×t =  <u>20</u>

  • 13. 6×3×t + 6×4×s = <u>144</u>

  • 14. 10×(10×x) + 10×y = <u>560</u>

  • 15. x·x·y = <u>150</u>

  • 16. r·s·t = <u>24</u>

  • 17. r·s·t·x·y = <u>720</u>

  • 18. 6·(2·x + 3·y) = <u>168</u>

<h3>Methods by which the above responses are arrived at:</h3>

Given; q = 1; r = 2; s = 3; t = 4; x = 5; y = 6

7. 2·(x + 4) + 10, by substituting x = 5, we have;

  • 2·(x + 4) + 10 = 2 × (5 + 4) + 10 = <u>28</u>

8. 3·(4·x) + 4

Substituting x = 5 gives;

  • 3·(4 × 5) + 4 = <u>64</u>

9. 10·(x·y)

Substituting x = 5, and y = 6 gives;

  • 10·(x·y) = 10 × (5 × 6) = <u>300</u>

10. 2·(x + y)

Substituting x = 5, and y = 6 gives;

  • 2·(x + y) = 2 × (5 + 6) =<u> 22</u>

11. 3·r·s + 4·x

Substituting r = 2, and s = 3, and x = 5, gives;

  • 3·r·s + 4·x = 3 × (2 × 3) + 4×(5) = <u>38</u>

12. 2·q + 2·x + 2×t  

Where, q = 1, x = 5, and t = 4, we get;

  • 2·q + 2·x + 2·t = 2 × 1 + 2 × (5 + 4) = <u>20</u>

13. 6·3·t + 6·4·s

Plugging in t = 4, and s = 3, we get;

  • 6·3·t + 6·4·s = 6 × (3×4 + 4×3) = <u>144</u>

14. 10·(10·x) + 10·y

x = 5, y = 6

  • 10·(10·x) + 10·y = 10 × (10×5) + 10 × 6 = <u>560</u>

15. x·x·y

x = 5, y = 6

Therefore;

  • x·x·y = 5 × (5 × 6) = <u>150</u>

16. r·s·t

r = 2, s = 3, t = 4

Therefore;

  • r·s·t = 2 × (3) × (4) = <u>24</u>

17. r·s·t·x·y

r = 2, s = 3, t = 4, x = 5, and y = 6

  • r·s·t·x·y = 2 × 3 × 4 × 5 × 6 = <u>720</u>

18. 6·(2·x + 3·y)

x = 5, y = 6

  • 6·(2·x + 3·y) = 6 × (2 × 5 + 3 × 6) =<u> 168</u>

<u />

<u />

<h3>Formatted Responses to the Questions;</h3>

Please find attached a representation of the above responses in the question format

Learn more about substitution property here:

brainly.com/question/11388301

brainly.com/question/11452422

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Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).
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Answer:

The arc length is \dfrac{21}{16}

Step-by-step explanation:

Given that,

The given curve between the specified points is

x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}

The points from (\dfrac{9}{16},1) to (\dfrac{9}{8},2)

We need to calculate the value of \dfrac{dx}{dy}

Using given equation

x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}

On differentiating w.r.to y

\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})

\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})

\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})

\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}

We need to calculate the arc length

Using formula of arc length

L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}

Put the value into the formula

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}

L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}

L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}

L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}

L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}

L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}

Put the limits

L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})

L=\dfrac{21}{16}

Hence, The arc length is \dfrac{21}{16}

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Answer:

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Consider the sequence:
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Answer:

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Step-by-step explanation:

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