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3 years ago

Given f(x)=4x-6 and g(x)=8x+4 findthe value of x for which f(x)=g(x)

Mathematics

1 answer:

lana [24]3 years ago

4 0

The value of "x" is $\frac{-5}{2}$

Solution:

Given f(x) = 4x - 6

g(x) = 8x + 4

To find: value of x when f(x) = g(x)

We can equate f(x) and g(x) as per given statement and solve for x

f(x) = g(x)

Substituting the values of f(x) and g(x) in above expression,

4x - 6 = 8x + 4

4x - 8x = 4 + 6

-4x = 10

On solving,

$x = \frac{10}{-4}\\\\x = \frac{-5}{2}$

Thus value of "x" is $\frac{-5}{2}$

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Answer:

The standard deviation of the sampling distribution of sample means would be 0.8186.

Step-by-step explanation:

We are given that

Mean of population=23.2 pounds

Standard deviation of population=6.6 pounds

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We have to find the standard deviation of the sampling distribution of sample means.

We know that standard deviation of the sampling distribution of sample means

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Using the formula

The standard deviation of the sampling distribution of sample means

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$=0.8186$

Hence, the standard deviation of the sampling distribution of sample means would be 0.8186.

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Step-by-step explanation:

The quadratic formula would be the following...

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Using this information and the information provided in the image we can plug in the quadratic equation values in the image to form the quadratic formula.

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2 years ago

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Step-by-step explanation:

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2 years ago

You stand a known distance from the base of the tree, measure the angle of elevation the top of the tree to be 15â—¦ , and then

gogolik [260]

Answer:

The maximum possible error of in measurement of the angle is $d\theta_1 =(14.36p)^o$

Step-by-step explanation:

From the question we are told that

The angle of elevation is $\theta_1 = 15 ^o = \frac{\pi}{12}$

The height of the tree is h

The distance from the base is D

h is mathematically represented as

$h = D tan \theta$ Note : this evaluated using SOHCAHTOA i,e

$tan\theta = \frac{h}{D}$

Generally for small angles the series approximation of $tan \theta \ is$

$tan \theta = \theta + \frac{\theta ^3 }{3}$

So given that $\theta = 15 \ which \ is \ small$

$h = D (\theta + \frac{\theta^3}{3} )$

$dh = D (1 + \theta^2) d\theta$

=> $\frac{dh}{h} = \frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta$

Now from the question the relative error of height should be at most

$\pm p$ %

=> $\frac{dh}{h} = \pm p$

=> $\frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta = \pm p$

=> $d\theta = \pm \frac{\theta + \frac{\theta^3}{3} }{1+ \theta ^2} * \ p$

So for $\theta_1$

$d\theta_1 = \pm \frac{\theta_1 + \frac{\theta^3_1 }{3} }{1+ \theta_1 ^2} * \ p$

substituting values

$d [\frac{\pi}{12} ] = \pm \frac{[\frac{\pi}{12} ] + \frac{[\frac{\pi}{12} ]^3 }{3} }{1+ [\frac{\pi}{12} ] ^2} * \ p$

=> $d\theta_1 = 0.25 p$

Converting to degree

$d\theta_1 = (0.25* 57.29) p$

$d\theta_1 =(14.36p)^o$

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3 years ago