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3 years ago

If f(x) = 2x - 4 and g(x) = x^2+3, find each value. 19. (f - g)(x) 20. (f • g)(x)

Mathematics

1 answer:

just olya [345]3 years ago

5 0

Step-by-step explanation:

$(f - g)(x) = f(x) - g(x) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = (2x - 4) - ( {x}^{2} + 3) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2x - 4 - {x}^{2} - 3 \\ \red{ \boxed{\therefore (f - g)(x) = - {x}^{2} + 2x - 7}}$

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A cable hangs between two poles of equal height and 35 feet apart. At a point on the ground directly under the cable and x feet

gayaneshka [121]

Answer:

293.38 pounds

Step-by-step explanation:

We are given that

Distance between poles=35 feet

$h(x)=10+0.1(x^{1.5})$

Weight of cable=10.4 per linear foot

We have to find the weight of the cable.

Differentiate w.r.t

$h'(x)=0.1(1.5)x^{0.5}=0.15x^{0.5}$

$s=2\int_{0}^{17.5}\sqrt{1+(h'(x))^2}dx$

$s=2\int_{0}^{17.5}\sqrt{1+(0.15x^{0.5})^2}dx$

$s=2\int_{0}^{17.5}\sqrt{1+0.0225x}dx$

Let $1+0.0225x=t$

$dx=\frac{1}{0.0225}dt$

$s=\frac{2}{0.0225}\int_{0}^{17.5}\sqrt{t}dt$

$s=\frac{2}{0.0225}\times\frac{2}{3}[t^{\frac{3}{2}}]^{17.5}_{0}$

$s=2\times \frac{2}{3\times0.0225}[(1+0.0255x)^{\frac{3}{2}]^{17.5}_{0}$

$s=\frac{4}{3\times 0.0225}((1+0.0225(17.5))^{\frac{3}{2}-1)$

s=28.21

Weight of cable= $28.21\times 10.4=293.38$ pound

8 0

3 years ago

Professor Halen teaches a College Mathematics class. The scores on the midterm exam are normally distributed with a mean of 72.3

lbvjy [14]

Answer:

14.63% probability that a student scores between 82 and 90

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 72.3, \sigma = 8.9$