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2 years ago

10

4x-8y=9 what is the slope of a line perpendicular to this line

1 answer:

LenaWriter [7]2 years ago

3 0

The slope of this line is 1/2

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2 years ago

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Divide.<br><br> 3/5 ÷ 1 1/4<br><br><br> A.) 1/4<br><br> B.) 12/25<br><br> C.) 3/4<br><br> D.) 2 2/5

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1 year ago

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3 years ago

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2 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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3 years ago