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3 years ago

Write the standard form of the equation of the circle that passes through the point

Mathematics

1 answer:

AnnZ [28]3 years ago

5 0

Answer:

The standard form of the equation of the circle is $x^2+y^2=1$ .

Step-by-step explanation:

A circle is the set of points in a plane that lie a fixed distance, called the radius, from any point, called the center.

The equation of a circle in standard form is

$(x-h)^2+(y-k)^2=r^2$

where r is the radius of the circle, and h, k are the coordinates of its center.

When the center of the circle coincides with the origin $h=k=0$ , so

$(x-0)^2+(y-0)^2=r^2\\x^2+y^2=r^2$

We are also told that the circle contains the point (0, 1), so we will use that information to find the radius r.

$0^2+1^2=r^2\\r^2=0^2+1^2\\r^2=1\\r=\sqrt{1}$

Therefore, the standard form of the equation of the circle is $x^2+y^2=1$ .

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Explain why 2a+3b=5ab

Ahat [919]

That's not correct. The terms 2a and 3b are not like terms, so we cannot combine them to get 5ab. We simply leave it as 2a+3b.

If you had 2a+3a, then it would simplify to 5a

Similarly, 2b+3b = 5b

Or you could have 2ab+3ab = 5ab

The key is that the variable portions must match up to be able to add them.

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3 years ago

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Write as a product: x^2+ax^2–y–ay+cx^2–cy Can you plz help 20 points for correct answer.

lana66690 [7]

download an app called photomath

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3 years ago

Prove it please answer only if you know

deff fn [24]

Part (c)

We'll use this identity

$\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)\\\\$

to say

$\sin(A+45) = \sin(A)\cos(45) + \cos(A)\sin(45)\\\\\sin(A+45) = \sin(A)\frac{\sqrt{2}}{2} + \cos(A)\frac{\sqrt{2}}{2}\\\\\sin(A+45) = \frac{\sqrt{2}}{2}(\sin(A)+\cos(A))\\\\$

Similarly,

$\sin(A-45) = \sin(A + (-45))\\\\\sin(A-45) = \sin(A)\cos(-45) + \cos(A)\sin(-45)\\\\\sin(A-45) = \sin(A)\cos(45) - \cos(A)\sin(45)\\\\\sin(A-45) = \sin(A)\frac{\sqrt{2}}{2} - \cos(A)\frac{\sqrt{2}}{2}\\\\\sin(A-45) = \frac{\sqrt{2}}{2}(\sin(A)-\cos(A))\\\\$

-------------------------

The key takeaways here are that

$\sin(A+45) = \frac{\sqrt{2}}{2}(\sin(A)+\cos(A))\\\\\sin(A-45) = \frac{\sqrt{2}}{2}(\sin(A)-\cos(A))\\\\$

Therefore,

$2\sin(A+45)*\sin(A-45) = 2*\frac{\sqrt{2}}{2}(\sin(A)+\cos(A))*\frac{\sqrt{2}}{2}(\sin(A)-\cos(A))\\\\2\sin(A+45)*\sin(A-45) = 2*\left(\frac{\sqrt{2}}{2}\right)^2\left(\sin^2(A)-\cos^2(A)\right)\\\\2\sin(A+45)*\sin(A-45) = 2*\frac{2}{4}\left(\sin^2(A)-\cos^2(A)\right)\\\\2\sin(A+45)*\sin(A-45) = \sin^2(A)-\cos^2(A)\\\\$

The identity is confirmed.

==========================================================

Part (d)

$\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)\\\\\sin(45+A) = \sin(45)\cos(A) + \cos(45)\sin(A)\\\\\sin(45+A) = \frac{\sqrt{2}}{2}\cos(A) + \frac{\sqrt{2}}{2}\sin(A)\\\\\sin(45+A) = \frac{\sqrt{2}}{2}(\cos(A)+\sin(A))\\\\$

Similarly,

$\sin(45-A) = \sin(45 + (-A))\\\\\sin(45-A) = \sin(45)\cos(-A) + \cos(45)\sin(-A)\\\\\sin(45-A) = \sin(45)\cos(A) - \cos(45)\sin(A)\\\\\sin(45-A) = \frac{\sqrt{2}}{2}\cos(A) - \frac{\sqrt{2}}{2}\sin(A)\\\\\sin(45-A) = \frac{\sqrt{2}}{2}(\cos(A)-\sin(A))\\\\$

-----------------

We'll square each equation

$\sin(45+A) = \frac{\sqrt{2}}{2}(\cos(A)+\sin(A))\\\\\sin^2(45+A) = \left(\frac{\sqrt{2}}{2}(\cos(A)+\sin(A))\right)^2\\\\\sin^2(45+A) = \frac{1}{2}\left(\cos^2(A)+2\sin(A)\cos(A)+\sin^2(A)\right)\\\\\sin^2(45+A) = \frac{1}{2}\cos^2(A)+\frac{1}{2}*2\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\\sin^2(45+A) = \frac{1}{2}\cos^2(A)+\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\$

and

$\sin(45-A) = \frac{\sqrt{2}}{2}(\cos(A)-\sin(A))\\\\\sin^2(45-A) = \left(\frac{\sqrt{2}}{2}(\cos(A)-\sin(A))\right)^2\\\\\sin^2(45-A) = \frac{1}{2}\left(\cos^2(A)-2\sin(A)\cos(A)+\sin^2(A)\right)\\\\\sin^2(45-A) = \frac{1}{2}\cos^2(A)-\frac{1}{2}*2\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\\sin^2(45-A) = \frac{1}{2}\cos^2(A)-\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\$

--------------------

Let's compare the results we got.

$\sin^2(45+A) = \frac{1}{2}\cos^2(A)+\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\\sin^2(45-A) = \frac{1}{2}\cos^2(A)-\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\$

Now if we add the terms straight down, we end up with $\sin^2(45+A)+\sin^2(45-A)$ on the left side

As for the right side, the sin(A)cos(A) terms cancel out since they add to 0.

Also note how $\frac{1}{2}\cos^2(A)+\frac{1}{2}\cos^2(A) = \cos^2(A)$ and similarly for the sin^2 terms as well.

The right hand side becomes $\cos^2(A)+\sin^2(A)$ but that's always equal to 1 (pythagorean trig identity)

This confirms that $\sin^2(45+A)+\sin^2(45-A) = 1$ is an identity

4 0

3 years ago

The owner of a construction company would like to know if his current work teams can build room additions quicker than the time

frez [133]

Answer:

The claim that the current work teams can build room additions quicker than the time allotted for by the contract has strong statistical evidence.

Step-by-step explanation:

We have to test the hypothesis to prove the claim that the work team can build room additions quicker than the time allotted for by the contract.

The null hypothesis is that the real time used is equal to the contract time. The alternative hypothesis is that the real time is less thant the allotted for by the contract.

$H_0: \mu=0\\\\H_1: \mu$

The significance level, as a storng evidence is needed, is α=0.01.

The estimated standard deviation is:

$s_M=\frac{s}{\sqrt{n}}=\frac{0.2}{\sqrt{16}} =0.2/4=0.05$

As the standard deviation is estimated, we use the t-statistic with (n-1)=15 degrees of freedom.

For a significance level of 0.01, right-tailed test, the critical value of t is t=2.603.

Then, we calculate the t-value for this sample:

$t=\frac{x-\mu}{s_M} =\frac{-0.32-0}{0.05}= -6.4$

As the t-statistic lies in the rejection region, the null hypothesis is rejected. The claim that the current work teams can build room additions quicker than the time allotted for by the contract has strong statistical evidence.

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3 years ago

if a person puts 1 cent in a piggy bank in the first day, 2 cents on the second day, 3 cents on the third day, and so on, how mu

MariettaO [177]

The answer would be: $8.20

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3 years ago

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