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3 years ago

Which ordered pair is a solution to the equation y=4x

Mathematics

1 answer:

Sedbober [7]3 years ago

6 0

The answer is going to be (1,3)
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Giải phương trình sau:<br> sinx=1/2

cupoosta [38]

Answer: $\begin{bmatrix}\mathrm{Radians:}\:&\:x=\frac{\pi }{6}+2\pi n,\:x=\frac{5\pi }{6}+2\pi n\:\\ \:\mathrm{Degrees:}&\:x=30^{\circ \:}+360^{\circ \:}n,\:x=150^{\circ \:}+360^{\circ \:}n\end{bmatrix}$

Step-by-step explanation:

$\sin \left(x\right)=\frac{1}{2}$

$x=\frac{\pi }{6}+2\pi n,\:x=\frac{5\pi }{6}+2\pi$

4 0

2 years ago

Read 2 more answers

Let F⃗ =2(x+y)i⃗ +8sin(y)j⃗ .

Alik [6]

Answer:

-42

Step-by-step explanation:

The objective is to find the line integral of $F$ around the perimeter of the rectangle with corners (4,0), (4,3), (−3,3), (−3,0), traversed in that order.

We will use <em>the Green's Theorem </em>to evaluate this integral. The rectangle is presented below.

We have that

$F(x,y) = 2(x+y)i + 8j \sin y = \langle 2(x+y), 8\sin y \rangle$

Therefore,

$P(x,y) = 2(x+y) \quad \wedge \quad Q(x,y) = 8\sin y$

Let's calculate the needed partial derivatives.

$P_y = \frac{\partial P}{\partial y} (x,y) = (2(x+y))'_y = 2\\Q_x =\frac{\partial Q}{\partial x} (x,y) = (8\sin y)'_x = 0$

Thus,

$Q_x -P_y = 0 -2 = - 2$

Now, by the Green's theorem, we have

$\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA = \int \limits_{-3}^{4} \int \limits_{0}^{3} (-2)\,dy\, dx \\ \\\phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-2y) \Big|_{0}^{3} \; dx\\ \phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-6)\; dx = -6x \Big|_{-3}^{4} = -42$

4 0

3 years ago

How many sides does the covered polygon have?

andrezito [222]

5 sides and 3 angles.

6 0

3 years ago

A line includes the points (0,2) and (1,6) What is the equation of the line in slope intercept form?

erica [24]

Answer:

y = 4x + 2

Step-by-step explanation:

$\frac{y^{2}-y^{1} }{x^{2} -x^{1} }$

$\frac{6-2}{1-0} = \frac{4}{1}$

y = 4x + 2

4 0

3 years ago

What amount of money will accumulate to $192,400 in 6 years and 11 months at the interest rate of 5.36% compounded semi-annually

MaRussiya [10]

We will have the following:

$\begin{gathered} 192400=A(1+\frac{0.0536}{2})^{(11/12)(2)}\Rightarrow A=\frac{192400}{(1+0.0536/2)^{(11/12)(2)}} \\ \\ \Rightarrow A=183293.7505...\Rightarrow A\approx18329375 \end{gathered}$

So, the original quantity was approximately $183 293.75.

3 0

1 year ago