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3 years ago

I need help on this to

Mathematics

1 answer:

vladimir1956 [14]3 years ago

6 0

13. 4cm is bigger

14. is equal

15. is equal

16. 4m is bigger

17. 150cm is bigger

18. 3km is bigger

19. is equal

20. 10km is bigger

21. 35mm is bigger

22. 55mm is bigger

23. 1km is bigger

24. is equal

25. 55mm is bigger

26. 2m is bigger

27. 8km is bigger

hope this helps

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14/100 = 0.14
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3 years ago

Suppose 8x + 16 ice cream cones

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Answer:

B. 152 + 7

Step-by-step explanation:

(8x + 16) + (7x - 9)

8x + 16 + 7x - 9

15x + 16 - 9

15x + 7

None of these options have this answer, so we see A and B:

A. x + 7

B. 152 + 7

It can't be A because we have 15x and A is x + 7

By process of elimination we have B. 152 + 7

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2 years ago

PLEASE HELP ME IM IN A BIG HURRY

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Part A:
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3 years ago

You are given the following equation.

saul85 [17]

Answer:

Step-by-step explanation:

Given the equation 4x²+ 49y² = 196

a) Differentiating implicitly with respect to y, we have;

$8x + 98y\frac{dy}{dx} = 0\\98y\frac{dy}{dx} = -8x\\49y\frac{dy}{dx} = -4x\\\frac{dy}{dx} = \frac{-4x}{49y}$

b) To solve the equation explicitly for y and differentiate to get dy/dx in terms of x,

First let is make y the subject of the formula from the equation;

If 4x²+ 49y² = 196

49y² = 196 - 4x²

$y^{2} = \frac{196}{49} - \frac{4x^{2} }{49} \\y = \sqrt{\frac{196}{49} - \frac{4x^{2} }{49} \\} \\$

Differentiating y with respect to x using the chain rule;

Let $u= \frac{196}{49} - \frac{4x^{2} }{49}$

$y = \sqrt{u} \\y =u^{1/2} \\$

$\frac{dy}{dx} = \frac{dy}{du} * \frac{du}{dx}$

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

$\frac{du}{dx} = 0 - \frac{8x}{49} \\\frac{du}{dx} =\frac{-8x}{49} \\\frac{dy}{dx} = \frac{1}{2} ( \frac{196}{49} - \frac{4x^{2} }{49})^{-1/2} * \frac{-8x}{49}\\\frac{dy}{dx} = \frac{1}{2} ( \frac{196-4x^{2} }{49})^{-1/2} * \frac{-8x}{49}\\\frac{dy}{dx} = \frac{1}{2} ( \sqrt{ \frac{49}{196-4x^{2} })} * \frac{-8x}{49}\\\frac{dy}{dx} = \frac{1}{2} *{ \frac{7}\sqrt {196-4x^{2} }} * \frac{-8x}{49}\\$

$\frac{dy}{dx} = \frac{-4x}{7\sqrt{196-4x^{2} } }$

c) From the solution of the implicit differentiation in (a)

$\frac{dy}{dx} = \frac{-4x}{49y}$

Substituting $y = \sqrt{\frac{196}{49} - \frac{4x^{2} }{49} \\$ into the equation to confirm the answer of (b) can be shown as follows

$\frac{dy}{dx} = \frac{-4x}{49\sqrt{\frac{196-4x^{2} }{49} } }\\\frac{dy}{dx} = \frac{-4x}{49\sqrt{196-4x^{2}}/7} }\\\\\frac{dy}{dx} = \frac{-4x}{7\sqrt{196-4x^{2}}}$

This shows that the answer in a and b are consistent.

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3 years ago

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3 years ago

Read 2 more answers

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