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2 years ago

If f(x)=x2+10sinx, show that there is a number c such that f(c)=1000.

1 answer:

7 0

F(x) is continuous for all x.

Pick a point and show that f(x) is either negative or positive. Pick another point and show that f(x) is negative, if positive, or positive, if negative.

At x = 30, f(30) - 1000 = 900 + 10sin(30) - 1000 ≤ 0
Now, show at another point f(x) - 1000 is positive, and hence, there would be root between 30 and such point.

Let's pick 40.
At x = 40, f(40) - 1000 = 1600 + 10sin(40) - 1000 ≥ 0

Since f(x) - 1000 is continuous, there lies a root between 30 and 40, and hence, 30 ≤ c ≤ 40

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$R^{2} + S^{2} = T^{2}$

Step-by-step explanation:

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2 years ago

Round 22360 to 1 significant figure

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This link to a video will help

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Please help!! I need to get this done! Im begging you, brainliest and points!

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Answer:

Solution; ( 1, 0 ), and ( - 3, 4 )

Step-by-step explanation:

See procedure below;

First let us make these two functions ( y = -x + 1 , y = x^2 + x - 2 ) equivalent;

- x + 1 = x^2 + x - 2,

-x - x + 1 + 2 - x^2 = 0,

- 2x + 3 - x^2 = 0,

x^2 + 2x - 3 = 0,

Now factor the simplified equation;

x^2 + 2x - 3 = 0,

( x^2 - x ) + ( 3x - 3 ) = 0,

x ( x - 1 ) + 3 ( x - 1 ) = 0,

( x - 1 )( x + 3 ) = 0,

And solve for x;

x - 1 = 0, and x + 3 = 0,

x = 1, and x = - 3

Now substitute this value of x into the two functions as to receive the y values for each x - value;

y = - ( 1 ) + 1, y = 0 for x = 1,

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Solution; ( 1, 0 ), and ( - 3, 4 )

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1 year ago

A print shop makes bumper stickers for election campaigns. If x stickers are ordered (where x < 10,000), then the price per s

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Answer:

P(x) = (0.049x - 0.0000015x²)

Step-by-step explanation:

price per sticker is 0.14 − 0.000002x dollars

total cost of producing the order is 0.091x − 0.0000005x² dollars.

P(x) = profit = Revenue - Cost

Let the number of units of stickers made be x

Revenue = (price per sticker) × (total units sold) = (0.14 − 0.000002x) × (x)

= (0.14x - 0.000002x²) dollars.

Cost of producing x units in the order = (0.091x − 0.0000005x²)

P(x) = 0.14x - 0.000002x² - (0.091x − 0.0000005x²) = 0.14x - 0.091x - 0.000002x² + 0.0000005x²

= (0.049x - 0.0000015x²)

P(x) = (0.049x - 0.0000015x²)

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2 years ago

A random sample of 35 undergraduate students who completed two years of college were asked to take a basic mathematics test. The

lisov135 [29]

Answer:

$(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96$

$(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96$

And the confidence interval would be given by:

$-1.96 \leq \mu_1 -\mu_2 \leq 7.96$

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

Step-by-step explanation:

For this case we have the following info given :

$\bar X_1= 75.1$ represent the sample mean for the scores of the undergraduate students

$s_1 = 12.8$ represent the standard deviation for the undergraduate students

$n_1 =35$ the sample size for the undergraduate

$\bar X_2= 72.1$ represent the sample mean for the scores of the high school students

$s_2 = 14.6$ represent the standard deviation for the high school students

$n_2 =50$ the sample size for the high school

The confidence interval for the true difference of means is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom are given by:

$df=n_1 +n_2 -2= 35+50-2=83$

The confidence level is 90% and the significance level is $\alpha=0.1$ and $\alpha/2 =0.05$ then the critical value would be:

$t_{\alpha/2}= 1.99$

And replacing the info we got:

$(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96$

$(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96$

And the confidence interval would be given by:

$-1.96 \leq \mu_1 -\mu_2 \leq 7.96$

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

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1 year ago