Ask question

Ask question

All categories

2 years ago

10

A rectangle has a width of six units and the length of 2X +8 units. If the area is 108 units squared, what is the value of X?

1 answer:

elena-s [515]2 years ago

5 0

Width of rectangle = 6 units

Length of rectangle = 2X+8 units

Area = 108 units square

Area = length * width

$108=6(2x+8)$

$108=12x+48$

$12x=60$

$x=5$ units

You might be interested in

Why do you think we have to separate items in a list using commas

stich3 [128]

Answer:

I think we have to use commas in lists because if we didn't it would be all together and it would be more hard to read.

Step-by-step explanation:

I hope this helps

8 0

2 years ago

Write the linear equation in slope-intercept form passing through the point (0, 3) and that has a slope of 2.

Yuri [45]

Answer:

y=2x+3

Step-by-step explanation:

5 0

2 years ago

In the given figure ABCD is a tripepizeum with AB||CD. If AO = x-1, CO = BO = x+1 and CD = x+4. Find the value of x.

Hitman42 [59]

$\longmapsto$ The value of "x" is 5.

$\large\underline{\sf{Solution-}}$

Given that,

A trapezium ABCD in which AB || CD such that

AO = x - 1

CO = BO = x + 1

OD = x + 4

Now,

$\rm In \: \triangle \: AOB \: and \: \triangle \: COD$

$\rm \: \angle \: AOB \: and \: \angle \: COD \: \: \{vertically \: opposite \: angles \}$

$\rm \: \angle \:ABO \: and \: \angle \: CDO \: \: \{alternate \: interior \: angles \}$

$\bf \: \triangle \: AOB \: \sim \: \triangle \: COD \: \: \: \{AA \: similarity \}$

$\bf\longmapsto\:\dfrac{AO}{CO} = \dfrac{BO}{DO}$

$\rm \longmapsto\:\dfrac{x - 1}{x + 1} = \dfrac{x + 1}{x + 4}$

$\rm \longmapsto\:(x - 1)(x + 4) = {(x + 1)}^{2}$

$\rm \longmapsto\: {x}^{2} - x + 4x - 4 = {x}^{2} + 1 + 2x$

$\rm \longmapsto\: 3x - 4 = 1 + 2x$

$\rm \longmapsto\: 3x - 2x= 1 +4$

$\bf\longmapsto \:x = 5$

7 0

2 years ago

What is the completely factored form of x4 + 8x2 – 9? (x + 1)(x – 1)(x + 3)(x + 3) (x+1)(x+1)(x+3){x^2+9} (x2 – 1)(x + 3)(x – 3)

Y_Kistochka [10]

Answer:

(x+1)(x-1)(x^2+9)

Step-by-step explanation:

5 0

2 years ago

Solve the simultaneous equation y-2x+1=0 and 4x^2+3y^2-2xy=7

Lapatulllka [165]

The first equation is $y-2x+1=0$

$y=2x-1$ (Equation 1)

The second equation is $4x^{2}+3y^{2}-2xy=7$ (Equation 2)

Putting the value of x from equation 1 in equation 2.

we get,

$4x^{2}+3(2x-1)^{2}-2x(2x-1)=7$

$4x^{2}+3(4x^{2}+1-4x)-4x^{2}+2x=7$

by simplifying the given equation,

$12x^{2}-10x-4=0$

$6x^{2}-5x-2=0$

Using discriminant formula,

$D=b^{2}-4ac$

$D=25-4 \times 6 \times -2 = 73$

Now the formula for solution 'x' of quadratic equation is given by:

$x=\frac{-b+\sqrt{D}}{2a} and x=\frac{-b-\sqrt{D}}{2a}$

$x=\frac{5+\sqrt{73}}{12} and x=\frac{5-\sqrt{73}}{12}$

Hence, these are the required solutions.

8 0

2 years ago