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AfilCa [17]
3 years ago
15

Can u answer this please

Mathematics
1 answer:
Svetradugi [14.3K]3 years ago
4 0

Answer:

C

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

(x - 4)² + (y - 3)² = 16 ← is in standard form → C

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Which is the best type of graph to show the number of boys and the number of girls who like soccer, basketball, or tennis?
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Number graph

Step-by-step explanation:

This way you can accurately count the number of each girl and boy  

6 0
3 years ago
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Paula, lohn, Kelsie, and Eart found 6 quarters, 3 dimes, and 2 pennies in the laundry machine. They decide they will share the m
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7 0
3 years ago
A terminal ray passes through (12,-5). Fine the value to all six trig functions
Harman [31]

Okay, here we have this:

Considering the provided angle, we are going to calculate the requested trigonometric functions, so we obtain the following:

So the first thing we will do is calculate the length of the hypotenuse, that is, the distance between the given point and the origin, then we have:

\begin{gathered} r=\sqrt{(12-0)^2+(-5-0)^2} \\ r=\sqrt{12^2+(-5)^2} \\ r=\sqrt{144+25} \\ r=\sqrt{169} \\ r=13 \end{gathered}

Now we proceed to find the value of each ratio:

sin\beta=\frac{y}{r}=\frac{-5}{13}cos\beta=\frac{x}{r}=\frac{12}{13}tan\beta=\frac{y}{x}=\frac{-5}{12}csc\beta=\frac{r}{y}=\frac{13}{-5}=-\frac{13}{5}sec\beta=\frac{r}{x}=\frac{13}{12}cot\beta=\frac{x}{y}=\frac{12}{-5}=-\frac{12}{5}

8 0
2 years ago
Solve the problem, calculate the line integral of f along h
Over [174]
The curve \mathcal H is parameterized by

\begin{cases}X(t)=R\cos t\\Y(t)=R\sin t\\Z(t)=Pt\end{cases}

so in the line integral, we have

\displaystyle\int_{\mathcal H}f(x,y,z)\,\mathrm ds=\int_{t=0}^{t=2\pi}f(X(t),Y(t),Z(t))\sqrt{\left(\frac{\mathrm dX}{\mathrm dt}\right)^2+\left(\frac{\mathrm dY}{\mathrm dt}\right)^2+\left(\frac{\mathrm dZ}{\mathrm dt}\right)^2}\,\mathrm dt
=\displaystyle\int_0^{2\pi}Y(t)^2\sqrt{(-R\sin t)^2+(R\cos t)^2+P^2}\,\mathrm dt
=\displaystyle\int_0^{2\pi}R^2\sin^2t\sqrt{R^2+P^2}\,\mathrm dt
=\displaystyle\frac{R^2\sqrt{R^2+P^2}}2\int_0^{2\pi}(1-\cos2t)\,\mathrm dt
=\pi R^2\sqrt{R^2+P^2}

You are mistaken in thinking that the gradient theorem applies here. Recall that for a scalar function f:\mathbb R^n\to\mathbb R, we have gradient \nabla f:\mathbb R^n\to\mathbb R^n. The theorem itself then says that the line integral of \nabla f(x,y,z)=\mathbf f(x,y,z) along a curve C parameterized by \mathbf r(t), where a\le t\le b, is given by

\displaystyle\int_C\mathbf f(x,y,z)\,\mathrm d\mathbf r=f(\mathbf r(b))-f(\mathbf r(a))

Specifically, in order for this theorem to even be considered in the first place, we would need to be integrating with respect to a vector field.

But this isn't the case: we're integrating f(x,y,z)=y^2, a scalar function.
7 0
3 years ago
Waise got 12, or 3/5 of her answers on the test right. How many questions were on the test?
zhannawk [14.2K]
There were 20 questions on the test because 3/5 x 4 = 12/20
hope this helped:)
4 0
4 years ago
Read 2 more answers
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