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4 years ago

What figure is described below? the points in space equidistant from the endpoints of PQ.

1 answer:

7 0

The question is slightly weird.

You have a piece of line, and its ends are 'P' and 'Q'.

All of the points on the paper that are the same distance
from 'P' and 'Q' are the points that form the line that bisects
segment-PQ and is perpendicular to it.
Other words: The line that is perpendicular to PQ at its mid-point.
Other words: The perpendicular bisector of segment PQ.

All of the points in space that are the same distance from
'P' and 'Q' are the points that form the plane perpendicular
to segment PQ at its mid-point.

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In the figures below, the cube-shaped box is 6 inches wide and the rectangular box is 10 inches long, 4 inches wide, and 4 inche

olganol [36]

The cube-shaped box has a volume of 216 cubic inches. The rectangular box has a volume of 160 cubic inches. Therefore, the answer is C) 56 cubic inches.

7 0

4 years ago

Find the perimeter of rhombus star

Degger [83]

Answer:

$4\sqrt{10}$

Step-by-step explanation:

Perimeter of the rhombus, STAR, is the sum of the length of all it's 4 sides.

The coordinates of its vertices are given as,

S(-1, 2)

T(2, 3)

A(3, 0)

R(0, -1)

Length of each side can be calculated using the distance formula given as $d = \sqrt{x_2 - x_1)^2 + (y_2 - y_1)^2}$

Find the length of each side ST, TA, AR, RS, using the above formula by plugging in the coordinate values (x, y) of each vertices.

$ST = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

S(-1, 2) => (x1, y1)

T(2, 3) => (x2, y2)

$ST = \sqrt{(2 -(-1))^2 + (3 - 2)^2}$

$ST = \sqrt{(3)^2 + (1)^2} = \sqrt{9 + 1} = \sqrt{10}$

$TA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

T(2, 3) => (x1, y1)

A(3, 0) => (x2, y2)

$TA = \sqrt{(3 - 2)^2 + (0 - 3)^2}$

$TA = \sqrt{(1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$

$AR = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

A(3, 0) => (x1, y1)

R(0, -1) => (x2, y2)

$AR = \sqrt{(0 - 3)^2 + (-1 - 0)^2}$

$AR = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$

$RS = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

R(0, -1) => (x1, y1)

S(-1, 2) => (x2, y2)

$RS = \sqrt{(-1 - 0)^2 + (2 -(-1))^2}$

$RS = \sqrt{(-1)^2 + (3)^2} = \sqrt{1 + 9} = \sqrt{10}$

$Perimeter = ST + TA + AR + RS$

$Perimeter = \sqrt{10} + \sqrt{10} + \sqrt{10} + \sqrt{10} = 4\sqrt{10}$

4 0

4 years ago

In ΔABC, ∠B measures 35° and the values of a and b are 19 and 11, respectively. Find the remaining measurements of the triangle,

Alexxx [7]

Answer:

(A)∠A = 82.2°,∠C = 62.8°, c = 17.1

Step-by-step explanation:

In Triangle ABC

∠B=35°

a=19

b=11

Using Law of SInes

$\dfrac{a}{\sin A} =\dfrac{b}{\sin B} \\\dfrac{19}{\sin A} =\dfrac{11}{\sin 35^\circ} \\11*\sin A=19*\sin 35^\circ\\\sin A=(19*\sin 35^\circ) \div 11\\A= \arcsin [(19*\sin 35^\circ) \div 11]\\A=82.2^\circ$

Now:

$\angle A+\angle B+\angle C=180^\circ\\35^\circ+82.2^\circ+\angle C=180^\circ\\\angle C=180^\circ-[35^\circ+82.2^\circ]\\\angle C=62.8^\circ$

Using Law of Sines

$\dfrac{c}{\sin C} =\dfrac{a}{\sin A} \\\dfrac{c}{\sin 62.8^\circ} =\dfrac{19}{\sin 82.2^\circ}\\c=\dfrac{19}{\sin 82.2^\circ}*\sin 62.8^\circ\\\\c=17.1$