Is 4m^2 - 3m = n linear or nonlinear??

Which method correctly solves the equation using the multiplication property of equality and the reciprocal of One-third? One-th

Airida [17]

Answer:

Step-by-step explanation:

Given the equation $\frac{1}{3}(x+9) = -12$

Step 1;

Expand the bracket at the right hand side of the equation to have:

$\frac{1}{3}x +\frac{1}{3}(9) = -12\\\frac{1}{3}x+3=-12\\subtracting\ 3\ from\ both\ sides\\\frac{1}{3}x+3-3=-12-3\\\frac{1}{3}x=-15\\$

Taking the reciprocal of both sides:

$\frac{3}{x} = \frac{-1}{15} \\ cross\ multiplying\\-x =45\\x=-45$

4 0

4 years ago

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Can you help me Round this ??

garri49 [273]

Answer:

It is 0.67

If the number after the number you want to round is bigger than 5 then the number would be added with 1

6 0

3 years ago

Derivative of <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%7B3x%7D%5E%7B2%7D%20-%202x%20-%201%20%7D%7B%20%7Bx%7D%5E%7B2

Anastaziya [24]

Answer:

$\displaystyle \frac{dy}{dx} = \frac{2x + 2}{x^3}$

Step-by-step explanation:

we would like to figure out the derivative of the following:

$\displaystyle \frac{ { 3x }^{2} - 2x - 1 }{ {x}^{2} }$

to do so, let,

$\displaystyle y = \frac{ { 3x }^{2} - 2x - 1 }{ {x}^{2} }$

By simplifying we acquire:

$\displaystyle y = 3 - \frac{2}{x} - \frac{1}{ {x}^{2} }$

use law of exponent which yields:

$\displaystyle y = 3 - 2 {x}^{ - 1} - { {x}^{ - 2} }$

take derivative in both sides:

$\displaystyle \frac{dy}{dx} = \frac{d}{dx} (3 - 2 {x}^{ - 1} - { {x}^{ - 2} } )$

use sum derivation rule which yields:

$\rm\displaystyle \frac{dy}{dx} = \frac{d}{dx} 3 - \frac{d}{dx} 2 {x}^{ - 1} - \frac{d}{dx} {x}^{ - 2}$

By constant derivation we acquire:

$\rm\displaystyle \frac{dy}{dx} = 0 - \frac{d}{dx} 2 {x}^{ - 1} - \frac{d}{dx} {x}^{ - 2}$

use exponent rule of derivation which yields:

$\rm\displaystyle \frac{dy}{dx} = 0 - ( - 2 {x}^{ - 1 -1} ) - ( - 2 {x}^{ - 2 - 1} )$

simplify exponent:

$\rm\displaystyle \frac{dy}{dx} = 0 - ( - 2 {x}^{ -2} ) - ( - 2 {x}^{ - 3} )$

two negatives make positive so,

$\displaystyle \frac{dy}{dx} = 2 {x}^{ -2} + 2 {x}^{ - 3}$

<h3>further simplification if needed:</h3>

by law of exponent we acquire:

$\displaystyle \frac{dy}{dx} = \frac{2 }{x^2}+ \frac{2}{x^3}$

simplify addition:

$\displaystyle \frac{dy}{dx} = \frac{2x + 2}{x^3}$

and we are done!

5 0

3 years ago

Addie bought lunch to go from a sandwich shop. She spent $12.00 for a $10.00 meal. What percent was the tip?

horsena [70]

Answer:

She gave a $2.00 tip witch would be %20 of her meal cost.

Step-by-step explanation:

</3 PureBeauty

4 0

3 years ago

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A Statistics professor has observed that for several years students score an average of 114 points out of 150 on the semester ex

kirill [66]

Answer:

Reject $H_0$ . The change is statistically significant. The software does appear to improve exam scores.

Step-by-step explanation:

We are given that a Statistics professor has observed that for several years students score an average of 114 points out of 150 on the semester exam. The software is expensive, and the salesman offers to let the professor use it for a semester to see if the scores on the final exam increase significantly.

In the trial course that used this software, 218 students scored an average of 117 points on the final with a standard deviation of 8.7 points.

Let $\mu$ = mean scores on the final exam.

SO, Null Hypothesis, $H_0$ : $\mu \leq$ 114 points {means that the mean scores on the final exam does not increases after using software}

Alternate Hypothesis, $H_A$ : $\mu$ > 114 points {means that the mean scores on the final exam increase significantly after using software}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average points = 117 points

s = sample standard deviation = 8.7 points

n = sample of students = 218

So, test statistics = $\frac{117-114}{\frac{8.7}{\sqrt{218} } }$ ~ $t_2_1_7$

= 5.091

Now, P-value of the test statistics is given by the following formula;

P-value = P( $t_2_1_7$ > 5.091) = Less than 0.05%

Since, in the question we are not given with the level of significance at which hypothesis can be tested, so we assume it to be 5%. Now at 5% significance level, the t table gives critical value of 1.645 at 217 degree of freedom for right-tailed test. Since our test statistics is higher than the critical value of t as 5.091 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the change is statistically significant. The software does appear to improve exam scores.

3 0

4 years ago