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3 years ago

Please answer the following ASAP

Mathematics

1 answer:

mr Goodwill [35]3 years ago

8 0

Answer:

Graphing: Graphing is the best method to use when introducing a new student to solving systems of two equations in two variables, because it gives them a visiual to recognize what they are looking for. Graphing is less exact and often takes more time than the other methods. I only recommend graphing to find a solution if the problem comes with a graph already drawn and the intersection appears to be on an exact coordinate.

Substitution: Substition gives that advantage of having an equation alrady written for the second variable when you find the first one. Substitution is best used when one (or both) of the equations is already solved for one of the variables. It also works well if one of the variables has a coefficient of 1.

Elimination: Elimination is the method that I use almost every time. If you are not sure which method to use, I recommend that you use elimination. Elimination is best used when both equations are in standard form (Ax + By = C). Elimination is also the best method to use if all of the variables have a coefficient other than 1.

You might be interested in

What is -5/12 - 4/14 + 6/34

andrew11 [14]

Answer:

4 1/2

Step-by-step explanation:

Look at this expression as given in the original problem; the numbers, properly typewritten, are -5 1/2, - 4 1/4, + 6 3/4.

We want to combine these three numbers into one.

To do this, we need the LCD; it is 4.

Thus, -5 1/2 is rewritten as -5 2/4.

Then we have

-5 2/4 - 4 1/4 + 6 3/4 = -5 -4 + 6 + 2/4 + 1/4 + 3/4.

This simplifies to: +3 + 6/4, or 3 + 1 + 2/4, or

4 1/2

5 0

3 years ago

If x is a rational number and y is the opposite of x, why do x and y have the same absolute value

grin007 [14]

Answer:

Absolute value is the distance away from zero, so if x is a rational number and y is opposite, they would both be the same distance away from zero, regardless of their sign.

Step-by-step explanation:

Absolute value is the distance of a number away from zero. For example, if x is equal to the rational number -2, the absolute value of -2 is just 2, given that both 2 and -2 are only a distance of 2 away from the number zero. Think of it as your walk or drive to school. If you live 4 miles from school, you drive a distance of 4 miles there and 4 miles back. That distance is not calculated as a positive 4 miles there and a negative 4 miles back, but rather just 4 miles there and 4 miles back, for a total distance of 8 miles. So, if x=4 and y=(-4), the absolute value of both would be 4.

8 0

2 years ago

Trouble finding arclength calc 2

kiruha [24]

Answer:

$S\approx1.1953$

Step-by-step explanation:

So we have the function:

$y=3-x^2$

And we want to find the arc-length from:

$0\leq x\leq \sqrt3/2$

By differentiating and substituting into the arc-length formula, we will acquire:

$\displaystyle S=\int\limits^\sqrt3/2}_0 {\sqrt{1+4x^2} \, dx$

To evaluate, we can use trigonometric substitution. First, notice that:

$\displaystyle S=\int\limits^\sqrt3/2}_0 {\sqrt{1+(2x)^2} \, dx$

Let's let y=2x. So:

$y=2x\\dy=2\,dx\\\frac{1}{2}\,dy=dx$

We also need to rewrite our bounds. So:

$y=2(\sqrt3/2)=\sqrt3\\y=2(0)=0$

So, substitute. Our integral is now:

$\displaystyle S=\frac{1}{2}\int\limits^\sqrt3}_0 {\sqrt{1+y^2} \, dy$

Let's multiply both sides by 2. So, our length S is:

$\displaystyle 2S=\int\limits^\sqrt3}_0 {\sqrt{1+y^2} \, dy$

Now, we can use trigonometric substitution.

Note that this is in the form a²+x². So, we will let:

$y=a\tan(\theta)$

Substitute 1 for a. So:

$y=\tan(\theta)$

Differentiate:

$y=\sec^2(\theta)\, d\theta$

Of course, we also need to change our bounds. So:

$\sqrt3=\tan(\theta), \theta=\pi/3\\0=\tan(\theta), \theta=0$

Substitute:

$\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{1+\tan^2(\theta)}\sec^2(\theta) \, d\theta$

The expression within the square root is equivalent to (Pythagorean Identity):

$\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{\sec^2(\theta)}\sec^2(\theta) \, d\theta$

Simplify:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta$

Now, we have to evaluate this integral. To do this, we can use integration by parts. So, let's let u=sec(θ) and dv=sec²(θ). Therefore:

$u=\sec(\theta)\\du=\sec(\theta)\tan(\theta)\, d\theta$

And:

$dv=\sec^2(\theta)\, d\theta\\v=\tan(\theta)$

Integration by parts:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\tan^2(\theta)\sec(\theta)} \, d\theta)$

Again, let's using the Pythagorean Identity, we can rewrite tan²(θ) as:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^2(\theta)-1)\sec(\theta)} \, d\theta)$

Distribute:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^3(\theta)-\sec(\theta)} \, d\theta)$

Now, let's make the single integral into two integrals. So:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta-\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)$

Distribute the negative:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta$

Notice that the integral in the first equation and the second integral in the second equation is the same. In other words, we can add the second integral in the second equation to the integral in the first equation. So:

$\displaystyle 2S= 2\int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta$

Divide the second and third equation by 2. So: $\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\frac{1}{2}(\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)$