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1 year ago

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What is the difference? StartFraction 2 x + 5 Over x squared minus 3 x EndFraction minus StartFraction 3 x + 5 Over x cubed minu

s 9 x EndFraction minus StartFraction x + 1 Over x squared minus 9 EndFraction StartFraction (x + 5) (x + 2) Over x cubed minus 9 x EndFraction StartFraction (x + 5) (x + 4) Over x cubed minus 9 x EndFraction StartFraction negative 2 x + 11 Over x cubed minus 12 x minus 9 EndFraction StartFraction 3 (x + 2) Over x squared minus 3 x EndFraction

1 answer:

lisov135 [29]1 year ago

3 0

Answer:

A. StartFraction (x + 5) (x + 2) Over x cubed minus 9 x EndFraction

Step-by-step explanation:

Given:

(2x + 5) / (x² - 3x) - (3x + 5) / (x³ - 9x) - (x + 1) / x² - 9

Factor the denominators

(2x + 5) / x(x - 3) - (3x + 5) / x(x - 3)(x + 3) - (x + 1) / (x - 3)(x + 3)

Lowest common multiple of the 3 fractions is x(x - 3)(x + 3)

= (2x+5)(x+3) - (3x + 5) - (x + 1)x / x(x - 3)(x + 3)

= (2x²+6x+5x+15) - (3x + 5) - (x² + x) / x(x - 3)(x + 3)

= 2x² + 11x + 15 - 3x - 5 - x² - x / x(x - 3)(x + 3)

= x² + 7x + 10 / x(x - 3)(x + 3)

Solve the numerator.

Solve the quadratic expression by finding two numbers whose product is 10 and sum is 7

The numbers are 5 and 2

= x² + 5x + 2x + 10 / x(x - 3)(x + 3)

= x(x + 5) + 2(x + 5) / x(x - 3)(x + 3)

= (x + 5)(x + 2) / x(x - 3)(x + 3)

A. StartFraction (x + 5) (x + 2) Over x cubed minus 9 x EndFraction

Recall,

x(x - 3)(x + 3) is a factor of x³ - 8x

A. StartFraction (x + 5) (x + 2) Over x cubed minus 9 x EndFraction

(x + 5)(x + 2) / x³ - 9x

B. StartFraction (x + 5) (x + 4) Over x cubed minus 9 x EndFraction

(x + 5)(x + 4) / x³ - 9x

C. StartFraction negative 2 x + 11 Over x cubed minus 12 x minus 9 EndFraction

2x + 11 / x³ - 12x - 9

D. StartFraction 3 (x + 2) Over x squared minus 3 x EndFraction

3(x + 2) / x² - 3x

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Answer:

Circumference: 64π

Ratio: 1 : 4

Measure of ∠xoy: π/2

Step-by-step explanation:

We are given an arc length of 16π. Since it's in terms of pi, we use the formula

S = rФ where r is the radius, and Ф is the measure of the angle in radians (in terms of pi)

We are given S = 16π and r = 32, plug those in and find Ф

16π = 32Ф

16π/32 = Ф

π/2 = Ф

This is the measure of the central angle.

The angle is π/2 radians. There are 2π radians in the circumference, so the circumference is 4 times the arc length created by the central angle. (There are 4 halves in 2) so the ratio of the arc length tothe circumference is 1 : 4

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C = 2π(32) = 64π

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1 year ago

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Solve for q. 22q+73 > 52q + 63

sammy [17]

Hey!

Let's write the problem.
$22q+73\ \textgreater \ 52q+63$
Subtract $73$ from both sides.
$22q+73-73\ \textgreater \ 52q+63-73$
$22q\ \textgreater \ 52q-10$
Subtract $52q$ from both sides.
$22q-52q\ \textgreater \ 52q-10-52q$
$-30q\ \textgreater \ -10$
Since we don't want negative, we will multiply both sides by $-1$ .
$\left(-30q\right)\left(-1\right)\ \textless \ \left(-10\right)\left(-1\right)$
$30q\ \textless \ 10$
Divide both sides by $30$ .
$\frac{30q}{30}\ \textless \ \frac{10}{30}$

Our final answer would be,
$q\ \textless \ \frac{1}{3}$

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An Article in the Journal of Sports Science (1987, Vol. 5, pp. 261-271) presents the results of an investigation of the hemoglob

ValentinkaMS [17]

Answer:

The 95% confidence interval for the population variance is $\left[0.219, \hspace{0.1cm} 0.807\right]\\\\$

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Step-by-step explanation:

To solve this problem, a confidence interval of $(1-\alpha) \times 100%$ for the population variance will be calculated.

$$$Sample variance: $S^2=(0.6152)^2$\\Sample size $n=20$\\Confidence level $(1-\alpha)\times100\%=95\%$\\$\alpha: \alpha=0.05$\\$\chi^2$ values (for a 95\% confidence and n-1 degree of freedom)\\$\chi^2_{\left (1-\frac{\alpha}{2};n-1\right )}=\chi^2_{(0.975;19)}=8.907\\$\chi^2_{\left (\frac{\alpha}{2};n-1\right )}=\chi^2_{(0.025;19)}=32.852\\\\$

Then, the $(1-\alpha) \times 100%$ confidence interval for the population variance is given by:

$\left [\frac{(n-1)S^2}{\chi^2_{\left (\frac{\alpha}{2};n-1\right )}}, \hspace{0.3cm}\frac{(n-1)S^2}{\chi^2_{\left (1-\frac{\alpha}{2};n-1\right )}} \right ]\\\\$ Thus, the 95% confidence interval for the population variance is: $\\\\\left [\frac{(19-1)(0.6152)^2}{32.852}, \hspace{0.1cm}\frac{(19-1)(0.6152)^2}{8.907} \right ]=\left[0.219, \hspace{0.1cm} 0.807\right]\\\\$

On other hand,

A confidence interval of $(1-\alpha) \times 100%$ for the population mean will be calculated

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2 years ago

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BaLLatris [955]

Answer:

Step-by-step explanation:

To find the HCF of 144 and 180

By using product of prime method

Firstly express 144 as a product of it prime and express 180 as a product of it prime

140=2×2×2×2×3×3

180=2×2×3×3×5

Common factor =2×2×3×3

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in term of m

m=36

13m-3

To find m

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465

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