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4 years ago

$3 for 6 bagels; $9 for 24 bagels.Determine if the rates or ratios are equivalent

Mathematics

2 answers:

castortr0y [4]4 years ago

8 0

Answer:

$3 for 6 bagels and $9 for 24 bagels are not equivalent.

Step-by-step explanation:

$3 for 6 bagels

i.e. 6 bangles cost $3

i.e. 6×3 bangles will cost $3×3

i.e. 18 bangles will cost $9

So, $3 for 6 bangles is equivalent to saying $9 for 18 bangles which is not equivalent to $9 for 24 bangles.

Hence, $3 for 6 bagels and $9 for 24 bagels are not equivalent.

Luba_88 [7]4 years ago

7 0

No these are not equivelent. 24 is 4x6 so that would mean that you would have to pay $12 for those bagels and because 9 is not 3x4, 12 is making these two ratios not equivelent.

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Veseljchak [2.6K]

Answer:

40 1/2 days

Step-by-step explanation:

Quantity of biscuits the Jones eats daily = 2/3 of a packet

Total packets of biscuits = 27 packets

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= Total packets of biscuits / daily intake of biscuits of the Jones

= 27 ÷ 2/3

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= 81 / 2

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Step-by-step explanation:

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

Find the point (,) on the curve =8 that is closest to the point (3,0). [To do this, first find the distance function between (,)

ELEN [110]

Question:

Find the point (,) on the curve $y = \sqrt x$ that is closest to the point (3,0).

[To do this, first find the distance function between (,) and (3,0) and minimize it.]

Answer:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

Step-by-step explanation:

$y = \sqrt x$ can be represented as: $(x,y)$

Substitute $\sqrt x$ for $y$

$(x,y) = (x,\sqrt x)$

So, next:

Calculate the distance between $(x,\sqrt x)$ and $(3,0)$

Distance is calculated as:

$d = \sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2}$

So:

$d = \sqrt{(x-3)^2 + (\sqrt x - 0)^2}$

$d = \sqrt{(x-3)^2 + (\sqrt x)^2}$

Evaluate all exponents

$d = \sqrt{x^2 - 6x +9 + x}$

Rewrite as:

$d = \sqrt{x^2 + x- 6x +9 }$

$d = \sqrt{x^2 - 5x +9 }$

Differentiate using chain rule:

Let

$u = x^2 - 5x +9$

$\frac{du}{dx} = 2x - 5$

So:

$d = \sqrt u$

$d = u^\frac{1}{2}$

$\frac{dd}{du} = \frac{1}{2}u^{-\frac{1}{2}}$

Chain Rule:

$d' = \frac{du}{dx} * \frac{dd}{du}$

$d' = (2x-5) * \frac{1}{2}u^{-\frac{1}{2}}$

$d' = (2x - 5) * \frac{1}{2u^{\frac{1}{2}}}$

$d' = \frac{2x - 5}{2\sqrt u}$

Substitute: $u = x^2 - 5x +9$

$d' = \frac{2x - 5}{2\sqrt{x^2 - 5x + 9}}$

Next, is to minimize (by equating d' to 0)

$\frac{2x - 5}{2\sqrt{x^2 - 5x + 9}} = 0$

Cross Multiply

$2x - 5 = 0$

Solve for x

$2x =5$

$x = \frac{5}{2}$

Substitute $x = \frac{5}{2}$ in $y = \sqrt x$

$y = \sqrt{\frac{5}{2}}$

Split

$y = \frac{\sqrt 5}{\sqrt 2}$

Rationalize

$y = \frac{\sqrt 5}{\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$y = \frac{\sqrt {10}}{\sqrt 4}$

$y = \frac{\sqrt {10}}{2}$

Hence:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

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3 years ago

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yawa3891 [41]

Answer:

Step-by-step explanation:

The zeros are the values of x for which y = 0

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= [5 ± 1] /2

= 2, 3

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3 years ago