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2 years ago

What is the equation of the circle with center (4, 4) that passes through the point (10, 14)? HELP MEEEEE PLEASE

Mathematics

2 answers:

hram777 [196]2 years ago

8 0

Answer:D

Step-by-step explanation:

Gemiola [76]2 years ago

5 0

Answer:

The equation of the circle can be written as:

$\left(x-4\right)^2+\left(y-4\right)^2=136$

Step-by-step explanation:

The general equation of a circle with center $(h,k)$ and radius $r$ is:

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

In our example, we know $\left(h,k\right)=\left(4,4\right)$ , as we just have to make sure we need determine $\:r^2$ .

$\left(x-4\right)^2+\left(y-4\right)^2=r^2\:\:$

As the circle passes through (10, 14), that pair of values for x and y must satisfy the equation. So we have:

$\left(10-4\right)^2+\left(14-4\right)^2=r^2$

$\mathrm{Switch\:sides}$

$r^2=\left(10-4\right)^2+\left(14-4\right)^2$

$r^2=6^2+10^2$

$r^2=36+100$

$r^2=136$

Thus the equation of the circle can be written as:

$\left(x-4\right)^2+\left(y-4\right)^2=136$

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Consider F and C below. F(x, y) = x4y5i + x5y4j, C: r(t) = t3 − 2t, t3 + 2t , 0 ≤ t ≤ 1 (a) Find a function f such that F = ∇f.

vovikov84 [41]

$\mathbf f(x,y)=x^4y^5\,\mathbf i+x^5y^4\,\mathbf j$

We want a scalar function $f(x,y)$ whose gradient is equivalent to the vector field. That means

$\dfrac{\partial f}{\partial x}=x^4y^5\implies f(x,y)=\dfrac{x^5y^5}5+g(y)$
$\dfrac{\partial f}{\partial y}=x^5y^4\implies x^5y^4=x^5y^4+\dfrac{\mathrm dg}{\mathrm dy}$
$\implies\dfrac{\mathrm dg}{\mathrm dy}=0\implies g(y)=C$

So (a)

$f(x,y)=\dfrac{x^5y^5}5+C$

which means (b)

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f(\mathbf r(1))-f(\mathbf r(0))=f(-1,3)-f(0,0)=-\dfrac{243}5-0=-\dfrac{243}5$

5 0

2 years ago

Simplify<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B4%7D%7B2%20%2B%20i%7D%20" id="TexFormula1" title=" \frac{4}{2 + i} "

skad [1K]

Hey! Apologies that I can’t make this look nice because I’m on mobile.

To simplify this, we can multiply the denominator by its conjugate. For a imaginary function (a+bi), its conjugate would be (a-bi), so we can apply that here to multiply both sides by (2-i)/(2-i). This works because it simplifies to one. Then, we have our fraction as 4*(2-i)/((2-i)(2+i))=(8-4i)/(4-i^2)=(8-4i)/5

Feel free to ask further questions, and have a great rest of your day!

6 0

3 years ago

F(x)=-2x+1; {-2,0,2,4,6} Find the range of the function for the given domain.

3241004551 [841]

Answer:

For the domain indicated, the range is {-11, -7, -3, 1, 5}

Step-by-step explanation:

Given the indicated domain of f(x) you have to replace the values in the function to find the range (the "y" values of the function).

f(-2) = 5

f(0) = 1

f(2) = -3

f(4) = -7

f(6) = -11.

So the range for this domain is {-11, -7, -3, 1, 5}

4 0

3 years ago

The Thomas Supply Company Inc. is a distributor of gas-powered generators. As with any business, the length of time customers ta

Dmitry [639]

Answer:

The answer is below

Step-by-step explanation:

When computing quartile and decile, the data must be arranged in ascending order.

Given the data points:

13 13 13 20 26 26 29 31 34 34 35 35 36 37 38 41 41 41 42 43 46 47 48 49 53 55 56 62 67 82

The numbers are arranged in ascending order. The total number of terms is 30.

$First \ quartile(Q_1)=\frac{1}{4}(n+1)^{th}\ term\\\\where\ n=number\ of\ terms. Hence:\\\\Q_1=\frac{1}{4}(30+1)=7.75^{th}\ term=average\ of\ 7th\ and\ 8th\ term=\frac{29+31}{2} =30\\\\Q_1=30\\\\Third \ quartile(Q_3)=\frac{3}{4}(n+1)^{th}\ term\\\\Q_3=\frac{3}{4}(30+1)=23.25^{th}\ term=average\ of\ 23rd\ and\ 24th\ term=\frac{48+49}{2} =48.5\\\\Q_3=48.5$

$Second \ decile(D_2)=\frac{2}{10}(n+1)^{th}\ term\\\\where\ n=number\ of\ terms. Hence:\\\\D_2=\frac{2}{10}(30+1)=6.2^{th}\ term=average\ of\ 6th\ and\ 7th\ term=\frac{26+29}{2} =27.5\\\\D_2=27.5\\\\\\Eight \ decile(D_8)=\frac{8}{10}(n+1)^{th}\ term=\frac{8}{10}(30+1)=24.8^{th}\ term\\=average\ of\ 24th\ and\ 25th\ term=\frac{49+53}{2} =51\\\\D_8=51$