Find the surface area and volume of the prism.<br> 13 cm<br> 28 cm<br> 5 cm<br> 12 cm

A rectangle has perimeter 28cm. Its area is 42sq.cm. Determine the dimensions of the rectangle. Include a diagram in your soluti

yawa3891 [41]

The dimensions of the rectangle are: b=9.65 cm and h=4.35 cm or b=4.35 cm and h= 9.65 cm.

<h3>Quadrilaterals</h3>

There are different types of quadrilaterals, for example, square, rectangle, rhombus, trapezoid, and parallelogram. Each type is defined accordingly to its length of sides and angles. For example, in a rectangle, the opposite sides are equal and parallel and their interior angles are equal to 90°.

The area of a rectangle is equal to base x height (A=bh). The perimeter of a geometric figure is the sum of its sides. Thus, for a rectangle, the perimeter is 2b+2h.

<h3>System of Linear Equations</h3>

System of linear equations is the given term math for two or more equations with the same variables. The solution of these equations represents the point at which the lines intersect.

The question gives:

Perimeter=28cm

Area= 42 cm²

If the perimeter is the sum of all sides of the rectangle, you have:

Perimeter= 2b+2h=28

If the area of the rectangle is 42cm², you have:

Area=bh=42 cm²

You can write a system of linear equations.

2b+2h=28 (1)

bh=42 (2)

From equation 2, you have $b=\frac{42}{h}$ . Then, you can replace it in equation 1.

$2*\frac{42}{h}+2h=28 \\ \\ 84+2h^2=28h\\ \\ 2h^2-28h+84=0 \; dividing\; by\; 2\\ \\ h^2-14h+42=0$

Now, you should solve the quadratic equation.

$h_{1,\:2}=\frac{-\left(-14\right)\pm \sqrt{\left(-14\right)^2-4\cdot \:1\cdot \:42}}{2\cdot \:1}$

$h_{1,\:2}=\frac{-\left(-14\right)\pm \:2\sqrt{7}}{2\cdot \:1}$

$h_1=\frac{-\left(-14\right)+2\sqrt{7}}{2\cdot \:1}=7+\sqrt{7}=9.65 cm$

$h_2=\frac{-\left(-14\right)-2\sqrt{7}}{2\cdot \:1}=7-\sqrt{7}=4.35cm$

If h1=9.65 cm, then $b_1=\frac{42}{9.65} =4.35 cm$ , and if h2=4.35 cm, then $b_1=\frac{42}{4.35} =9.65 cm$ .

Read more about the quadratic function here:

brainly.com/question/1497716

#SPJ1

8 0

2 years ago

If t=0 in 1980, find the value of A. Round answer to the nearest million.

balu736 [363]

P=Ae^kt

225 = 210 * e^(k*(1990-1980)

225/210=e^10k ln(225/210)=10k

k=ln(225/210)/10=0.0069

P = 210*e^(0.0069t)

for 2000 ===> t = 2000-1980=20

P = 210*e^(0.0069*20)

P=241.0749=241

4 0

3 years ago

Solve for p. 21 + p = 44 HELP THIS IS URGENT

Scorpion4ik [409]

P= 41-21
The final result is P= 20

4 0

3 years ago

Read 2 more answers

I WILL GIVE YOU A CROWN

katrin2010 [14]

Answer:

486 in³

Step-by-step explanation:

Area of base = l*b = 81sq. inches

h = 6 inches

Volume of the box = lbh = 81 * 6 = 486 cubic inches

6 0

3 years ago

You have 108 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the River

Licemer1 [7]

Answer:

The length and width of the plot that will maximize the area of the rectangular plot are 54 ft and 27 ft respectively.

Step-by-step explanation:

Given that,

The length of fencing of the rectangular plot is = 108 ft.

Let the longer side of the rectangular plot be x which is also the side along the river side and the width of the rectangular plot be y.

Since the fence along the river does not need.

So the total perimeter of the rectangle is =2(x+y) -x

=2x+2y-y

=x+2y

So,

x+2y =108

⇒x=108 -2y

Then the area of the rectangle plot is A = xy

A=xy

⇒A= (108-2y)y

⇒ A = 108y-2y²

A = 108y-2y²

Differentiating with respect to x

A'= 108 -4y

Again differentiating with respect to x

A''= -4

For maximum or minimum, A'=0

108 -4y=0

⇒4y=108

$\Rightarrow y=\frac{108}{4}$

⇒y=27.

$A''|_{y=27}=-4$

Since at y= 27, A''<0

So, at y=27 ft , the area of the rectangular plot maximum.

Then x= (108-2.27)

=54 ft.

The length and width of the plot that will maximize the area of the rectangular plot are 54 ft and 27 ft respectively.

3 0

3 years ago

Find the surface area and volume of the prism. 13 cm 28 cm 5 cm 12 cm