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3 years ago

7

Please help explain how to do this

2 answers:

dexar [7]3 years ago

8 0

Answer:

Any number that isn't 3.

Step-by-step explanation:

Any other number should work, so I'll just show you what happens if you use 3.

2(3x - 2) = 6x + 5

6x - 4 = 6x + 5

-4 ≠ 5

As you are aware, that would make it no solution hence why it can't be used.

ELEN [110]3 years ago

4 0

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On your paper, construct a rectangle on a coordinate plane that satisfies these criteria.

german

Answer:

$A = (5,3)$

$B = (-5,3)$

$C = (-5,-8)$

$D = (5,-8)$

Step-by-step explanation:

Required

$Perimeter = 42$

Construct a rectangle whose perimeter is 42 units and satisfies the given conditions.

First, name the rectangle ABCD.

Such that:

$A = (x_1,y_1)$

$B = (x_2,y_2)$

$C = (x_3,y_3)$

$D = (x_4,y_4)$

For the rectangle to be either horizontal or vertical, then:

$y_1 = y_2$ and $y_3 = y_4$

We have that:

$Perimeter = 42$

Replace perimeter with its formula

$2(AB + BC) = 42$

Divide both sides by 2

$AB + BC = 21$

This implies that, the distance between adjacent sides (through the edges) must be equal to 21

Having said that: a set of coordinates that satisfy the given conditions are:

$A = (5,3)$ -- First quadrant

$B = (-5,3)$ -- Second quadrant

$C = (-5,-8)$ -- Third quadrant

$D = (5,-8)$ -- Fourth quadrant

The above quadrants satisfy the condition:

$y_1 = y_2$ and $y_3 = y_4$

<u>HOW TO KNOW THE PERIMETER IS 42</u>

To do this, we simply calculate the distance between the edges and add them up

<u>Distance is calculated as:</u>

<u></u> $D = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$ <u></u>

<u></u>

<u>For AB</u>

$A = (5,3)$

$B = (-5,3)$

$D_1 = \sqrt{(5 - (-5))^2 + (3 - 3)^2}= \sqrt{(10)^2 + (0)^2} = \sqrt{100} = 10$

<u>For BC</u>

$B = (-5,3)$

$C = (-5,-8)$

$D_2 = \sqrt{(-5 - (-5))^2 + (3 - (-8))^2}= \sqrt{(0)^2 + (11)^2} = \sqrt{121} = 11$

<u>For CD</u>

$C = (-5,-8)$

$D = (5,-8)$

$D_3 = \sqrt{(-5 -5)^2 + (-8 - (-8))^2}= \sqrt{(-10)^2 + (0)^2} = \sqrt{100} = 10$

<u>For DA</u>

$D = (5,-8)$

$A = (5,3)$

$D_4 = \sqrt{(5 -5)^2 + (-8 -3)^2}= \sqrt{(0)^2 + (11)^2} = \sqrt{121} = 11$

So, the perimeter is:

$P = D_1 + D_2 + D_3 + D_4$

$P = 10 + 11 + 10 +11$

$P = 42$

See attachment for rectangle

8 0

3 years ago

Lori creates the following design for a T-shirt.

kakasveta [241]

<u>Solution-</u>

From the figure,

AE = 2.4

EB = 2.8

BC = 11.7

Area of rectangle 1 = 8.68 sq.in

$\Rightarrow FH \times HI=8.68$

$\Rightarrow EB \times HI=8.68$ (∵ sides of the rectangle 2)

$\Rightarrow 2.8 \times HI=8.68$

$\Rightarrow HI=3.1$

Area of Triangle 1 = 6.48 sq.in

$\Rightarrow \frac{1}{2}\times AE \times EG= 6.48$

$\Rightarrow \frac{1}{2}\times AE \times (EF+FG)= 6.48$

$\Rightarrow \frac{1}{2}\times AE \times (EF+HI)= 6.48$ (∵ sides of the rectangle 1)

$\Rightarrow EF+3.1= 5.4$

$\Rightarrow EF=2.3$

$\Rightarrow BH=2.3$ (∵ sides of the rectangle 2)

$BC = BH+HI+IC$

$\Rightarrow 11.7= 2.3+3.1+IC$

$\Rightarrow IC=6.3$

The area of Rectangle 2,

$=EB\times BH =2.8\times 2.3=6.44\ sq.in$

The area of Triangle 2,

$\frac{1}{2}\times GI \times IC=\frac{1}{2}\times EB \times IC=\frac{1}{2}\times 2.8 \times 6.3=8.82\ sq.in$

The area of the whole figure = Area of Triangle 1 + Area of rectangle 1 + Area of Triangle 2 + Area of rectangle 2

= 6.48+8.68+8.82+6.44=30.42 sq.in

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Answer:

Step-by-step explanation:

total price was reduced by $ 7

let x = the total price........ x - 7

he bought 10.....and it cost him $ 130

10(x - 7) = 130

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10x = 200

x = 200/10

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