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ZanzabumX [31]
2 years ago
9

What is the slope of the line through ( -2.1) and (2, -5)

Mathematics
1 answer:
nordsb [41]2 years ago
6 0

Answer:

ltxitxifi DC ffg curxurxutx

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ON THE SMARTBOARD, THE
Dafna11 [192]

answer

width = 98 cm

length = 105 cm

set up equations

the perimeter of a rectangle = 2W + 2L

p = 2W + 2L

the perimeter is also 406

406 = 2W + 2L

we know that the length of the rectangle is 7 cm longer than the width so

L = W + 7

substitute equations and solve for width

plug L into p = 2W + 2L to get

406 = 2W + 2(W + 7) distribute the 2 to the terms in the parentheses

406 = 2W + 2W + 14

406 = 4W + 14

392 = 4W

W = 98

substitute equations and solve for length

now that we know the width, we can substitute W in L = W + 7 to find length

L = W + 7

L = 98 + 7

L = 105

final dimensions

width = 98 cm

length = 105 cm

5 0
3 years ago
I need it ungently it’s due in 20 mins
Ipatiy [6.2K]

The answer is c you must look at key words i learned my key words

5 0
3 years ago
I need help with this problem. Given: BC = 10 inches AC = √50 inches m∠CBD = 60° m∠CAD = 90° Calculate the exact area of the sha
Nikitich [7]

Answer:

Area of the shaded region = 23.33 in²

Step-by-step explanation:

Area of a sector = \frac{\theta}{360}(\pi r^{2})

Where θ = Central angle subtended by an arc

r = radius of the circle

Area of the sector BCD = \frac{60}{360}(\pi) (10^{2})

                                       = 52.36 in²

Area of equilateral triangle BCD = \frac{\sqrt{3} }{4}(\text{Side})^2

                                                      = \frac{\sqrt{3} }{4}(10)^2

                                                      = 25\sqrt{3} in²

                                                      = 43.30 in²

Area of the shaded portion in ΔBCD = 52.36 - 43.3

                                                             = 9.06 in²

Area of sector CAD = \frac{90}{360}(\pi)(\sqrt{50})^2

                                 = 39.27 in²

Area of right triangle CAD = \frac{1}{2}(\text{Base})(\text{Height})

                                            = \frac{1}{2}(\text{AC})(\text{AD})

                                            = \frac{1}{2}(\sqrt{50})(\sqrt{50})

                                            = 25 in²

Area of the shaded part in the ΔACD = 39.27 - 25

                                                                         = 14.27 in²

Area of the shaded part of the figure = 9.06 + 14.27

                                                                = 23.33 in²

8 0
3 years ago
5 POINTS, 5 STARS, BRAINLIEST, AND THANKS
Aneli [31]
No cause it’s a curved line so it’s not direct
6 0
2 years ago
Name the transformations of the next equation y=3(x+6) - 5
klemol [59]

Answer:

vertical stretch

Step-by-step explanation:

8 0
2 years ago
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