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3 years ago

Help me please i need a 67 becasue i cant play games tommorwo if i get 64

Mathematics

2 answers:

liubo4ka [24]3 years ago

8 0

Step-by-step explanation:

3 ÷ 5 is 0.6

0.6 is the answer

Genrish500 [490]3 years ago

6 0

Answer:

$\frac{3}{5}$ or $0.6$ depending on if they want it as a fraction or a decimal.

Step-by-step explanation:

You would do 3 divided by 5 to get 0.6.

You would just do 3/5 to get the fraction.

Hope I helped

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What is the area of square UVWX?<br>Area =

guapka [62]

Answer:

Area = 64

Step-by-step explanation:

area of a square = $side^{2}$

side length = 8

$8^{2}$ = 64

8 0

3 years ago

Solve the triangle that has a=4.6, B=19°, A=92° (picture provided)

kolezko [41]

Answer:

Option b

Step-by-step explanation:

To solve this problem use the law of the sines.

We have 2 angles of the triangle and one of the sides.

$a = 4.6\\B = 19\°\\A = 92\°\\C = 180 -A - B\\C = 180 - 92 - 19\\C = 69\°$

The law of the sines is:

$\frac{sin(A)}{a} = \frac{sin(B)}{b} = \frac{sin(C)}{c}$

Then:

$\frac{sin(92)}{4.6} = \frac{sin(19)}{b}\\\\b = \frac{sin(19)}{\frac{sin(92)}{4.6}}\\\\b = 1.5$

$\frac{sin(B)}{b} = \frac{sin(C)}{c}\\\\\frac{sin(19)}{1.5} = \frac{sin(69)}{c}\\\\c = \frac{sin(69)}{\frac{sin(19)}{1.5}}\\\\c = 4.30$

8 0

3 years ago

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

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