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3 years ago

-4d - 5 = -1 answer for me pls

Mathematics

1 answer:

PolarNik [594]3 years ago

4 0

$\\ \sf\longmapsto -4d-5=-1$

Cancel - from both sides

$\\ \sf\longmapsto -(4d+5)=-(1)$

$\\ \sf\longmapsto 4d+5=1$

$\\ \sf\longmapsto 4d=1-5$

$\\ \sf\longmapsto 4d=-4$

$\\ \sf\longmapsto d=\dfrac{-4}{4}$

$\\ \sf\longmapsto d=-1$

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Answer:

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Step-by-step explanation:

Given that,

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Using given equation

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

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

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Using formula of arc length

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

Put the value into the formula

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

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

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

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

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

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

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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3 years ago

How many triangles can be constructed with sides measuring 5 m, 16 m, and 5 m?

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3 years ago

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Answer:

Step-by-step explanation:

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$\mathbf{|R|= 2^{10}}$

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or, ㏒ ( $0.88^{b}$ ) < ㏒ ( $\frac{1}{5.2}$ )

or, b× ㏒ 0.88 < ㏒ ( $\frac{1}{5.2}$ )

or, $\frac{log (\frac{1}{5.2}) }{log (0.88)}$ < b

or, 12.89 < b

Hence for the equation to have value less than 100, b must be greater than 12.89.

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4 years ago

Which value of x makes the equation below true? 7 + x = 84 A. 12 B. 77 C. 83 D. 91

garri49 [273]

Answer:

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Step-by-step explanation:

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