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andrew-mc [135]

2 years ago

8

A square rotated about its center by 360º maps onto itself at: a) 1 B) 2 C) 3 D) 4 different angles of rotation. You can reflect

a square onto itself across A) 2 B) 4 C) 8 D) 16 different lines of reflection.

2 answers:

grandymaker [24]2 years ago

4 0

Answer: A square rotated about its center by 360º maps onto itself at D) 4 different angles of rotation. You can reflect a square onto itself across B) 4 different lines of reflection.

Step-by-step explanation:

A square is a geometric figure which has all its four sides equal and all its interior angles are right angles( $90^{\circ}$ ) .

Therefore, it can be rotated about its center by $360^{\circ}$ .

It maps onto itself at 4 different angles of rotation ( at every $90^{\circ} angle$ ).

We can reflect a square onto itself across 4 different lines of reflection (2 across the non-parallel sides and 2 across the vertices of the square).

NeTakaya2 years ago

3 0

Answer:

For plato: Blank one is four and blank two is also four.

:)

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The table shows different quantities of flour and the number of cookies that can be made with each quantity of flour. If the rel

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3/4 = 6/x....3/4 = 6/8...notice that proportions are nothing but equivalent fractions

x = 8

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Ralph bought a computer monitor with an area of 384 square inches. The length of the monitor is six times the quantity of five l

Alenkasestr [34]

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2 years ago

Solve the equation a. 0.76, –0.76 c. 3.87 b. 4.98 d. 1.53, –1.53 Please select the best answer from the choices provided A B C D

lord [1]

You didn't write the equation you like to be solved. It is clear though, from the options you gave, that there are two equations you want to be solved simultaneously. I will give an example to illustrate this, and you may apply the process in solving your problem.

Step-by-step explanation:

Let

x + y = 7 ....................................(1)

x - 2y = -2...................................(2)

SOLVING USING THE SUBSTITUTION METHOD.

From (1), make y the subject.

y = 7 - x ....................................(3)

Substitute the value of y in (3) into (2)

x - 2(7 - x) = -2

x - 14 + 2x = -2

x + 2x = -2 + 14

3x = 12

Divide both sides by 3

x = 12/3 = 4

Now use x = 4 in (3)

y = 7 - 4 = 3

Therefore, ( x, y) = (4, 3)

SOLVING USING THE ELIMINATION METHOD.

x + y = 7 ....................................(1)

x - 2y = -2...................................(2)

First, let us eliminate x by subtracting (2) from (1)

(x + y) - (x - 2y) = 7 - (-2)

y + 2y = 7 + 2

3y = 9

Divide both sides by 3

y = 9/3 = 3

To eliminate y, first multiply (1) by 2, and add the result to (1)

2 × (1): 2x + 2y = 14 ........................(3)

................x - 2y = -2.........................(2)

____________________

...............3x = 12

............... x = 12/3 = 4

Therefore (x, y) = (4, 3)

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2 years ago

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The cost of tyre has increased by 20% to £78. work out the original price

tiny-mole [99]

The original price was 65

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2 years ago

Evaluate the line integral by the two following methods. xy dx + x2y3 dy C is counterclockwise around the triangle with vertices

nadezda [96]

Answer:

a)

$\frac{2}{3}$

b)

$\frac{2}{3}$

Step-by-step explanation:

a) The first part requires that we use line integral to evaluate directly.

The line integral is

$\int_C xydx + {x}^{2} {y}^{3} dy$

where C is counterclockwise around the triangle with vertices (0, 0), (1, 0), and (1, 2)

The boundary of integration is shown in the attachment.

Our first line integral is

$L_1 = \int_ {(0,0)}^{(1,0)} xydx + {x}^{2} {y}^{3} dy$

The equation of this line is y=0, x varies from 0 to 1.

When we substitute y=0 every becomes zero.

$\therefore \: L_1 =0$

Our second line integral is

$L_2 = \int_ {(1,0)}^{(1,2)} xydx + {x}^{2} {y}^{3} dy$

The equation of this line is:

$x = 0 \implies \: dx = 0$

y varies from 1 to 2.

We substitute the boundary and the values to get:

$L_2 = \int_ {1}^{2}1 \cdot y(0) + {1}^{2} \cdot \: {y}^{3} dy$

$L_2 = \int_ {1}^2 {y}^{3} dy = \frac{8}{3}$

The 3rd line integral is:

$L_3 = \int_ {(1,2)}^{(0,0)} xydx + {x}^{2} {y}^{3} dy$

The equation of this line is

$y = 2x \implies \: dy = 2dx$

x varies from 0 to 1.

We substitute to get:

$L_3 = \int_ {1}^{0} x \cdot \: 2xdx + {x}^{2} {(2x)}^{3}(2 dx)$

$L_3 = \int_ {1}^{0} 8 {x}^{5} + 2 {x}^{2} dx = - 2$

The value of the line integral is

$L = L_1 + L_2 + L_3$

$L = 0 + \frac{8}{3} + - 2 = \frac{2}{3}$

b) The second part requires the use of Green's Theorem to evaluate:

$\int_C xydx + {x}^{2} {y}^{3} dy$

Since C is a closed curve with counterclockwise orientation, we can apply the Green's Theorem.

This is given by:

$\int_C \: Pdx +Q \: dy = \int \int_ R \: Q_y - P_x \: dA$

$\int_C \: xydx + {x}^{2} {y}^{3} \: dy = \int \int_ R \: 3 {x}^{2} {y}^{2} - y \: dA$

We choose our region of integration parallel to the y-axis.

$\int_C \: xydx + {x}^{2} {y}^{3} \: dy = \int_ 0^{1} \int_ 0^{2x} \: 3 {x}^{2} {y}^{2} - y \: dydx$

$\int_C \: xydx + {x}^{2} {y}^{3} \: dy = \int_ 0^{1} \: {x}^{2} {y}^{3} - \frac{1}{2} {y}^{2} |_ 0^{2x} dx$

$\int_C \: xydx + {x}^{2} {y}^{3} \: dy = \int_ 0^{1} \: 8{x}^{5} - 2 {x}^{2} dx = \frac{2}{3}$

8 0

2 years ago