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3 years ago

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A bag contains white marbles and yellow marbles, 16 in total. The number of white marbles is 5 less than 2 times the number of y

ellow marbles. How many white marbles are there?

1 answer:

bija089 [108]3 years ago

8 0

Answer:

9 White marbles are present in the bag.

Step-by-step explanation:

Let the number of White marbles present in the bag be X and number of Yellow marbles present be Y .

Then according to the question ,

X + Y = 16 -(1) (As total marbles are 16)

Also,

X = 2Y - 5 -(2)

From Equation 1 , X = 16 - Y , Comparing with Equation 2 , we get,

16 - Y = 2Y - 5

3Y = 21.

Y = 7.

So, X = 16 - Y = 16 - 7 = 9.

Thus, 9 White marbles are present in the bag.

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Soloha48 [4]

Answer:

$f(216) \approx 6.0093$

Step-by-step explanation:

Given

$\sqrt[3]{217}$

Required

Solve

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$f(x + \triangle x) \approx f(x) +\triangle x \cdot f'(x)$

Take:

$x = 216; \triangle x= 1$

So:

$f(x) = \sqrt[3]{x}$

Substitute 216 for x

$f(x) = \sqrt[3]{216}$

$f(x) = 6$

So, we have:

$f(x + \triangle x) \approx f(x) +\triangle x \cdot f'(x)$

$f(215 + 1) \approx 6 +1 \cdot f'(x)$

$f(216) \approx 6 +1 \cdot f'(x)$

To calculate f'(x);

We have:

$f(x) = \sqrt[3]{x}$

Rewrite as:

$f(x) = x^\frac{1}{3}$

Differentiate

$f'(x) = \frac{1}{3}x^{\frac{1}{3} - 1}$

Split

$f'(x) = \frac{1}{3} \cdot \frac{x^\frac{1}{3}}{x}$

$f'(x) = \frac{x^\frac{1}{3}}{3x}$

Substitute 216 for x

$f'(216) = \frac{216^\frac{1}{3}}{3*216}$

$f'(216) = \frac{6}{648}$

$f'(216) = \frac{3}{324}$

So:

$f(216) \approx 6 +1 \cdot f'(x)$

$f(216) \approx 6 +1 \cdot \frac{3}{324}$

$f(216) \approx 6 + \frac{3}{324}$

$f(216) \approx 6 + 0.0093$

$f(216) \approx 6.0093$

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<u>Step 1: Define points</u>

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