M = 128° and m = 76°. What is m∠ABC?

Which of the following represents the set of even whole numbers greater than 10

romanna [79]

Need more information. You said which of the following and didn't put any questions under it

3 0

3 years ago

A statistics instructor likes to assign the material as reading in the textbook and then give a pretest before covering the mate

klio [65]

Answer:

b. The student's scores on the posttest would have a smaller standard deviation.

Step-by-step explanation:

The first test is taken before the material is covered in class so we expect the standard deviation to be high because not everyone's scores would be lying close to the mean. Equal number of students mastered most, some or almost none of the material from reading the textbook based on the pretest result. this means the data is varying, so the standard deviation is large.

Whereas, after the teacher has taught the material and given the homework, they must have understood most of the material. The test they take after teaching as a post test. The results of the post test would have a smaller standard deviation as most of the students would have scored good. Hence, the student's scores on the posttest would have a smaller standard deviation.

8 0

4 years ago

Can someone thoroughly explain this implicit differentiation with a trig function. No matter how many times I try to solve this,

Anton [14]

Answer:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}$

Step-by-step explanation:

So we have the equation:

$\tan(x-y)=\frac{y}{8+x^2}$

And we want to find dy/dx.

So, let's take the derivative of both sides:

$\frac{d}{dx}[\tan(x-y)]=\frac{d}{dx}[\frac{y}{8+x^2}]$

Let's do each side individually.

Left Side:

We have:

$\frac{d}{dx}[\tan(x-y)]$

We can use the chain rule, where:

$(u(v(x))'=u'(v(x))\cdot v'(x)$

Let u(x) be tan(x). Then v(x) is (x-y). Remember that d/dx(tan(x)) is sec²(x). So:

$=\sec^2(x-y)\cdot (\frac{d}{dx}[x-y])$

Differentiate x like normally. Implicitly differentiate for y. This yields:

$=\sec^2(x-y)(1-y')$

Distribute:

$=\sec^2(x-y)-y'\sec^2(x-y)$

And that is our left side.

Right Side:

We have:

$\frac{d}{dx}[\frac{y}{8+x^2}]$

We can use the quotient rule, where:

$\frac{d}{dx}[f/g]=\frac{f'g-fg'}{g^2}$

f is y. g is (8+x²). So:

$=\frac{\frac{d}{dx}[y](8+x^2)-(y)\frac{d}{dx}(8+x^2)}{(8+x^2)^2}$

Differentiate:

$=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}$

And that is our right side.

So, our entire equation is:

$\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}$

To find dy/dx, we have to solve for y'. Let's multiply both sides by the denominator on the right. So:

$((8+x^2)^2)\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}((8+x^2)^2)$

The right side cancels. Let's distribute the left:

$\sec^2(x-y)(8+x^2)^2-y'\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy$

Now, let's move all the y'-terms to one side. Add our second term from our left equation to the right. So:

$\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy+y'\sec^2(x-y)(8+x^2)^2$

Move -2xy to the left. So:

$\sec^2(x-y)(8+x^2)^2+2xy=y'(8+x^2)+y'\sec^2(x-y)(8+x^2)^2$

Factor out a y' from the right:

$\sec^2(x-y)(8+x^2)^2+2xy=y'((8+x^2)+\sec^2(x-y)(8+x^2)^2)$

Divide. Therefore, dy/dx is:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)+\sec^2(x-y)(8+x^2)^2}$

We can factor out a (8+x²) from the denominator. So:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}$

And we're done!

8 0

3 years ago

– 2y – 10 = x<br> - 10x - y = -14

fredd [130]

Answer:

hey there guys ^^

Step-by-step explanation:

5 0

3 years ago

How do I find the slope of the line through (–9, –10) and (–2, –5)?

storchak [24]

Answer:
slope of line is 5/7

Explanation:
The slope of the line can be calculated using the following rule:
slope = $\frac{y2-y1}{x2-x1}$

The given points are:
(-9,-10) which represent (x1,y1)
(-2,-5) which represent (x2,y2)

Substitute with the points in the above equation to get the slope as follows:
slope = $\frac{-5-(-10)}{-2-(-9)}$ = 5/7

Hope this helps :)

4 0

4 years ago

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