All categories

2 years ago

Use graphing to find the solutions to the system of equations.

Mathematics

1 answer:

Galina-37 [17]2 years ago

3 0

In order to utilize the graph, first you have to distinguish which graph accurately pertains to the two functions.

This can be done by rewriting the equations in the form y = mx + b which can be graphed with ease; where m is the slope and b is the y intercept.

-x^2 + y = 1
y = x^2 + 1

So this will be a basic y = x^2 parabola where the center intercepts on the y axis at (0, 1)

-x + y = 2
y = x +2

So this will be a basic y = x linear where the y intercept is on the y axis at (0, 2)

The choice which depicts these two graphs correctly is the first choice. The method to find the solutions to the system of equations by using the graph is by determining the x coordinate of the points where the two graphed equations intersect.

You might be interested in

Help Me, Will give 100 points.

Bad White [126]

Answer:

baka

Step-by-step explanation:

6 0

2 years ago

I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

4 0

2 years ago

Read 2 more answers

Please please please please help me someone

Shtirlitz [24]

Answer:

That incorrect

Step-by-step explanation:

Yes

3 0

2 years ago

HELPPPP! Which expression is equivalent to 7Vx^2/5Vy^2

baherus [9]

Pretty sure it’s the first answer

4 0

2 years ago

Read 2 more answers

A=student is female

Aleks04 [339]

<span>For "The probability a business major is female" - you're looking for the probability of being female. That the person is a business major is already given. So, P(A|B)

</span>For "The probability a female student is majoring in business" - you're looking for the probability of being majoring in business. That the person is a female is already given. So, P(B|A)

4 0

3 years ago