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3 years ago

Parallelogram BCDE with vertices B(2, -3), C(6,-4), D(7.-6), and E(3.-5):

Mathematics

1 answer:

Klio2033 [76]3 years ago

5 0

Rotation 270 counterclockwise about origin is easy, swap the x and y then reverse the sign of the y
so (x, y) rotates to (y, -x)
Sadly to change the centre we need to translate the centre to the origin, rotate and the translate back.
In this case that means adding -1, 2 to x,y respectively, rotating, adding 1, -2 to x,y respectively

B(2, 3) translates to (1, 5) which rotates to (5, -1) and translates back to B’(6, -3)
C(6, -4) translates to (5, -2) which rotates to (-2, -5) and translates back to C’(-1, -7)
D(7, -6) translates to (6, -4) which rotates to (-4, -6) and translates back to (-3, -8)
E(3, -5) translates to (2, -3) which rotates to (-3, -2) and translates back to (-2, -4)

The translation is much easier simply moving each point 8 to the right
B’(2, 11), C’(6, 4), D’(7, 2) and E’(3, 3)

Using the grid will everything clear!

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What is the sum of an infinite geometric series if the first term is 81 and the common ratio is 2⁄3?

Marrrta [24]

the sum to infinity of a geometric series = $\frac{a}{1-r}$

where a is the first term and r the common ratio

= $\frac{81}{\frac{1}{3} }$ = 81 × 3 = 243

4 0

4 years ago

Scott will rent a car for the weekend. He can choose one of two plans. the first plan has an initial fee of $55 and costs an add

Firlakuza [10]

Answer:

This problem is about linear equations. We assume Dale drive X miles, and the total cost is $Y, then we can get:

Plan I: Y=38+0.11X

Plan II: Y=49+0.07X

When both plans cost the same, 38+0.11X=49+0.07X. We will get X = 275miles, and Y=$68.25

Step-by-step explanation:

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3 years ago

In an isosceles trapezoid the length of a diagonal is 25 cm and the length of an altitude is 15 cm. Find the area of the trapezo

andreev551 [17]

Answer:

The area of the trapezoid is $525\ cm^{2}$

Step-by-step explanation:

we know that

The area of a isosceles trapezoid is equal to the area of two isosceles right triangles plus the area of a rectangle

step 1

<em>Find the area of the isosceles right triangle</em>

Remember that

In a isosceles right triangle the height is equal to the base of the triangle

we have

$h=15\ cm$

$b=15\ cm$

The area is equal to

$A=\frac{1}{2}(b)(h)$

substitute the values

$A=\frac{1}{2}(15)(15)=112.5\ cm^{2}$

step 2

Find the area of the rectangle

The area of the rectangle is equal to

$A=LW$

we have

$W=15\ cm$ -----> is the height of the trapezoid

$d=25\ cm$ -----> the diagonal of the rectangle

Applying the Pythagoras Theorem

$25^{2}=L^{2}+15^{2}\\L^{2}=25^{2}-15^{2} \\ L^{2} =400\\L=20\ cm$

The area of the rectangle is

$A=(20)(15)=300\ cm^{2}$

step 3

Find the area of the trapezoid

$A=2(112.5\ cm^{2})+300\ cm^{2}=525\ cm^{2}$

6 0

4 years ago

Read 2 more answers

The area of the rectangular garden shown is 19.5 square meters. How many meters of fencing are needed to go around the garden?

Elenna [48]

Answer:

17.66m

Step-by-step explanation:

Let the area of the rectangular garden be L²

A = L²

19.5 = L²

L = √19.5

L = 4.416

Perimeter = 4L

Perimeter = 4(4.416)

Perimeter of the garden =17.66m

Hence the number of meters of fencing that are needed to go around the garden is 17.66m

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2 years ago