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4 years ago

Find the measure of angle Z

Mathematics

1 answer:

ira [324]4 years ago

8 0

Answer:

∠Z ≈ 12°

Step-by-step explanation:

Using the Sine Rule in ΔXYZ

$\frac{x}{sinx}$ = $\frac{z}{sinz}$ , that is

$\frac{48}{sin96}$ = $\frac{10}{sinz}$ ( cross- multiply )

48 × sinZ = 10 × sin96° ( divide both sides by 48 )

sinZ = $\frac{10sin96}{48}$

Z = $sin^{-1}$ ( $\frac{10sin96}{48}$ ) ≈ 12°

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solve the equation using inverse operations check your solutions in your final answer include all of your work 5 - 2x^2 = -15

Ronch [10]

Answer:

x = ± $\sqrt{10}$

Step-by-step explanation:

Given

5 - 2x² = - 15 ( subtract 5 from both sides )

- 2x² = - 20 ( divide both sides by - 2 )

x² = 10 ( take the square root of both sides )

x = ± $\sqrt{10}$

7 0

3 years ago

Read 2 more answers

FIND THE EQUATION OF THE ELLIPSE WITH A CENTER AT (2, 2), VERTICES AT (-3,

Ivahew [28]

Answer:

Step-by-step explanation:

The standard equation of an ellipse centered at the point (h,k) is

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} = 1$

where a is the distance from the center to one of the vertex. We have the relation $c= \sqrt[]{a^2-b^2}$ where c is the distance from one of the focus to the center.

The distance between one vertex and the center is 5. So a=5. The distance from one focue to the center is 3. Then c =3. So we have that $b^2 = a^2-c^2 = 16$

so the equation is

$\frac{(x-2)^2}{25}+\frac{(y-2)^2}{16} = 1$

5 0

3 years ago

The coordinates of polygon ABCD are A(-4,-1), B(-2,3), C(2,2), and D(4,-3). Use these coordinates to complete the sentences belo

shepuryov [24]

Answer:

The perimeter of polygon ABCD, to the nearest thousandth units is

22.227 units

The perimeter of polygon ABCD', to the nearest thousandth, would be 20.980 units

The area of polygon ABCD' is 19.5 units²

Step-by-step explanation:

The coordinates forming the polygon are

A (-4,-1), B(-2,3), C(2,2), and D(4,-3)

The perimeter then is given by the sum of the length of the sides as follows;

Length of line between two X and Y points distance, xi and yj apart is

length XY = $\sqrt{x^2+y^2}$

Therefore, the length between points AB is

length AB = $\sqrt{(-4 - (-2))^2+(-1-3)^2}$

= $\sqrt{(-4 +2)^2+(-4)^2} = \sqrt{-2^2+-4^2} = \sqrt{20}$ = 4.472 units

Similarly, length BC is given by

Length BC = $\sqrt{(-4)^2+1^2} = \sqrt{17}$ = 4.123 units

Length CD = $\sqrt{(-2)^2+5^2} = \sqrt{29}$ = 5.385 units

Length DA = $\sqrt{(8)^2+(-2)^2} = \sqrt{68}$ = 8.246 units

The perimeter is equal to;

length AB + Length BC + Length CD + Length DA

= 4.472 + 4.123 + 5.385 + 8.246 = 22.227 units

If the point D is moved up 2 units and left 1 unit we have

D' = x-1, y+2 where x and y are the coordinates of point D

D(4, -3) → D'((4-1), (-3+2)) = D'(3, -1)

The length D'A = $\sqrt{(7)^2+(0)^2} = \sqrt{49}$ =7 units

The perimeter of polygon ABCD'

length AB + Length BC + Length CD + Length D'A

= 4.472 + 4.123 + 5.385 + 7 = 20.980 units.

The area is given by the determinant of the 3 by 3 matrix using Cramer's Rule as follows

$S_{\bigtriangleup}$ = $(\frac{1}{2})|x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2|$

= $(\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|$

Triangle ABC we have

A(-4,-1)

B(-2,3)

C(2,2)

$S_{{\bigtriangleup}ABC}$ = $(\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|$

$=(\frac{1}{2})|(-4)(3-2)+(-2)(2-(-1)) +2((-1)-3)|$

= $=(\frac{1}{2})|-4-6 -8|= 9 units^2$

Triangle ACD'

A(-4,-1)

C(2,2)

D'(3, -1)

$S_{{\bigtriangleup}ACD'}$ = $(\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|$

$=(\frac{1}{2})|(-4)(2+1)+(2)(-1+1) +3((-1)-2)| = \frac{21}{2}$ = 10.5 units²

Area of polygon = $S_{{\bigtriangleup}ABCD'}$ = $S_{{\bigtriangleup}ABC}$ + $S_{{\bigtriangleup}ACD'}$ = (9 + 10.5) units²

Area of polygon ABCD' = 19.5 units².

4 0

3 years ago

There are 16 boys and 12 girls in mr..harts science class. He wants to form as many groups as possible for a project. All the gr

ser-zykov [4K]

Answer:

Step-by-step explanation:there are only 12 girls so that is the only amount of groups to be formed

5 0

3 years ago

A particle moves along line segments from the origin to the points (2, 0, 0), (2, 3, 1), (0, 3, 1), and back to the origin under

sweet [91]

Answer: Work done by the particle is 78 N.

Step-by-step explanation:

If C is the path the particle follows, then work done is $\int_{C}^{}F.dx$ .

According to the question the force $F=z^{2}i+5xyj+4y^2k$ and all four points are in the plane $z=\frac{1}{4}y$ .

therefore, if S is the at surface with boundary C, so that S is the portion of the plane $z=\frac{1}{4}y$ over the rectangle D=[0,2]\times [0,4].

Now,

$curlF=\begin{vmatrix}i &j &k \\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ z^{2}&5xy &4y^{2} \end{vmatrix}=8yi+2zj+5yk$

By the Stoke's theorem:

$\int_{C}^{}F.dx=\iint_{s}^{}curlF.dS\\\\=\iint_{D}^{}[-8y(0)-2z\left ( \frac{1}{4} \right )+5y]dA\\\\=\iint_{D}^{}\left ( -\frac{1}{2}z+5y \right )dA$

$=\iint_{D}^{}\left ( \frac{39}{8}y \right )dA\, \, \, \, \, \, \, \, \, Since\, \, z=\frac{1}{4}y\\\\=\int_{0}^{2}\int_{0}^{4}\frac{39}{8}y\, \, \, dy\, dx\\\\$

$\int_{0}^{2}\left [ \frac{39}{16}y^2 \right ]_{0}^{4}dx\\\\=\int_{0}^{2}39\, \, dx\\\\=39\times 2\\\\=78 \, N$

8 0

4 years ago