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3 years ago

Estimate 6 8 9 ( 1 9 10 ) by rounding to the nearest whole number. The estimate is the actual product because

Mathematics

2 answers:

siniylev [52]3 years ago

4 0

Answer:

$6\frac{8}{9} (1\frac{9}{10}) = 13$

Step-by-step explanation:

Given

$6\frac{8}{9} (1\frac{9}{10})$

Required

Estimate

To do this; we start by converting both fractions to improper fraction

$6\frac{8}{9} (1\frac{9}{10})$

$\frac{62}{9} (1\frac{9}{10})$

$\frac{62}{9} (\frac{19}{10})$

Open Bracket

$\frac{62}{9} * \frac{19}{10}$

Combine to form a single fraction

$\frac{62 * 19}{9 * 10}$

Multiply numerators and denominators

$\frac{1178}{90}$

Convert fraction to decimal

$13.0888888889$

Approximate to nearest whole number

$13$

Hence, $6\frac{8}{9} (1\frac{9}{10}) = 13$

viktelen [127]3 years ago

3 0

Answer:

First box: 14

Second box: greater than

Third box: both factors were rounded up.

Step-by-step explanation:

I JUST DID IT AND GOT IT RIGHT!

HOPE IT HELPS!

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3 years ago

If x = a cosθ and y = b sinθ , find second derivative

Olin [163]

I'm guessing the second derivative is for <em>y</em> with respect to <em>x</em>, i.e.

$\dfrac{\mathrm d^2y}{\mathrm dx^2}$

Compute the first derivative. By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta$

$x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta$

Now compute the second derivative. Notice that $\frac{\mathrm dy}{\mathrm dx}$ is a function of $\theta$ ; so denote it by $f(\theta)$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$