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3 years ago

-8+ 8b = 16 8b+8=16 -8=-8 8b=8

Mathematics

1 answer:

e-lub [12.9K]3 years ago

7 0

$b = 3 \\ b = 1 \\ 0 \\ b = 1$

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The length of a rectangle is twice the width. An equation that models the perimeter of the rectangle is 2w+4w=36 where w is the

drek231 [11]

Answer: 12

Step-by-step explanation:

Let w = the width of the rectangle

Let 2w = the length

P = 2l + 2w

36 = 2(2w) + 2w

36 = 4w + 2w

36 = 6w

w = 6

l = 2(6)

l = 12

4 0

3 years ago

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A market vendor sells pawpaws and guavas. In a week the number of pawpaws sold and the number of guavas sold were in the ratio 3

EastWind [94]

Total awpaws and guavas sold in a week = 328

Let the number of pawpas and guavas be x

ATQ

3x + 5x = 328

8x = 328

x = 328/8

x = 41

Pawapas in a week = 123

Guavas in a week = 205

Must click thanks and mark brainliest

3 0

3 years ago

This question has three parts. Answer the parts in order.

nalin [4]

Answer:

The area of the smallest section is $A_{1}=100yd^{2}$

The area of the largest section is $A_{2}=625yd^{2}$

The area of the remaining section is $A_{3}=250yd^{2}$

Step-by-step explanation:

Please see the picture below.

1. First we are going to name the side of the larger square as x.

As the third section shares a side with the larger square and the four sides of a square are equal, we have the following:

- Area of the first section:

$A_{1}=10yd*10yd$

$A_{1}=100yd^{2}$

- Area of the second section:

$A_{2}=x^{2}$ (Eq.1)

- Area of the third section:

$A_{3}=width*length$

$A_{3}=10yd*x$ (Eq.2)

2. The problem says that the total area of the enclosed field is 975 square yards, and looking at the picture below, we have:

$A_{1}+A_{2}+A_{3}=975yd^{2}$

Replacing values:

$100+x^{2}+10x=975$

Solving for x:

$x^{2}+10x-875=0$

$x=\frac{-10+\sqrt{100+(4*875)}}{2}$

$x=\frac{-10+\sqrt{3600}}{2}$

$x=\frac{-10+60}{2}$

$x=25$

3. Replacing the value of x in Eq.1 and Eq.2:

- From Eq.1:

$A_{2}=25^{2}$

$A_{2}=625yd^{2}$

- From Eq.2:

$A_{3}=10*25$

$A_{3}=250yd^{2}$

3 0

2 years ago

The coordinates of the vertices of △RST are R(−3,1), S(−1,4), and T(3,1) .

Ghella [55]

<span>△RST is a right triangle because RS¯¯ is perpendicular to RT¯¯ .</span>

3 0

3 years ago

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Graph y=x+1/x² and y=1/x for positive values of x.

AlekseyPX

Answer:

Graphic is showed in the figure below

Step-by-step explanation:

To graph the equations given, let's do a table for positive values of x, and, by replacing it in the equation, let's calculate the value of y. Knowing the coordinate points (x,y) we can build the graphic.

1 2 1

2 2.25 0.2

3 3.11 0.33

4 4.06 0.25

When x->0 both equations -> ∞, because lim(1/x) x->0 = ∞

The graphic is showed below. In red there is y = 1 + 1/x² and in blue y = 1/x

6 0

3 years ago