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3 years ago

If f(x)=x^1/2-x and g(x)=2x^3-x^1/2-x, find f(x)-g(x)

Mathematics

2 answers:

inn [45]3 years ago

6 0

X^1/2 - x
-2x^3 - x^1/2 - x

-2x^3 - 2x
i think this is it

bogdanovich [222]3 years ago

3 0

The answer is d. It the one with the 1/2

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Please explain and ANSWER this question for BRAINLIEST!

pogonyaev

Answer:

$155 in^{2}$

Step-by-step explanation:

Hope this work helps

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2 years ago

1. Amanda weighs 48 pounds (lbs). The physician ordered a drug that is delivered in the proportion of milligrams (mg) of medicat

allsm [11]

48 pounds/1 * 1 kg/2.2 pounds=<span>21.82 kg
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3 years ago

If (x+3)÷ 3 = 9 what is x

Sveta_85 [38]

Answer:

x = 24

Step-by-step explanation:

9*3 = 27

27 - 3 = 24

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7 0

3 years ago

Read 2 more answers

Find the equation of the tangent to the circle x2 + y2 = 109 at point (-10,3)

konstantin123 [22]

Answer:

10x - 3y = -109 .

Step-by-step explanation:

x^2 + y^2 = 109

Implicit Differentiation:

2x + 2y.y' = 0

y' = -2x/2y = -x/y

So at the point (-10,3) the gradient of the tangent is -(-10)/3 = 10/3.

Equation of the tangent:

y - y1 = m(x - x1)

y - 3 = 10/3(x + 10)

y - 3 = 10/3 x + 100/3

3y - 9 = 10x + 100

-109 + 3y = 10x

10x - 3y = -109

4 0

3 years ago

A company receives shipments of a component used in the manufacture of a component for a high-end acoustic speaker system. When

Tanya [424]

Answer:

$ME= 2.33*\sqrt{\frac{0.096*(1-0.096)}{250}}= 0.0434$

Step-by-step explanation:

For this case we have a sample size of n = 250 units and in this sample they found that 24 units failed one or more of the tests.

We are interested in the proportion of units that fail to meet the company's specifications, and we can estimate this with:

$\hat p = \frac{24}{250}= 0.096$

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The confidence interval for a proportion is given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 98% confidence interval the value of $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.33$

And the margin of error would be:

$ME= 2.33*\sqrt{\frac{0.096*(1-0.096)}{250}}= 0.0434$

4 0

4 years ago