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3 years ago

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the length of a rectangle is 5 less than twice the width.if the perimeter of the rectangle is 146,find the area of the rectangle

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1 answer:

raketka [301]3 years ago

7 0

Answer:

The area of the rectangle is 1222 units²

Step-by-step explanation:

The formula of the perimeter of a rectangle is P = 2(L + W), where L is its length and W is its width

The formula of the area of a rectangle is A = L × W

∵ The length of a rectangle is 5 less than twice the width

- Assume that the width of the rectangle is x units and multiply

x by 2 and subtract 5 from the product to find its length

∴ W = x

∴ L = 2x - 5

- Use the formula of the perimeter above to find its perimeter

∵ P = 2(2x - 5 + x)

∴ P = 2(3x - 5)

- Multiply the bracket by 2

∴ P = 6x - 10

∵ The perimeter of the rectangle is 146 units

∴ P = 146

- Equate the two expression of P

∴ 6x - 10 = 146

- Add 10 to both sides

∴ 6x = 156

- Divide both sides by 6

∴ x = 26

Substitute the value of x in W and L expressions

∴ W = 26 units

∴ L = 2(26) - 5 = 52 - 5

∴ L = 47 units

Now use the formula of the area to find the area of the rectangle

∵ A = 47 × 26

∴ A = 1222 units²

∴ The area of the rectangle is 1222 units²

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Consider the parabola given by the equation: f(x) = 4x² - 6x - 8 Find the following for this parabola: A) The vertex: Preview B)

jeyben [28]

Answer:

The vertex: $(\frac{3}{4},-\frac{41}{4} )$

The vertical intercept is: $y=-8$

The coordinates of the two intercepts of the parabola are $(\frac{3+\sqrt{41} }{4} , 0)$ and $(\frac{3-\sqrt{41} }{4} , 0)$

Step-by-step explanation:

To find the vertex of the parabola $4x^2-6x-8$ you need to:

1. Find the coefficients <em>a</em>, <em>b</em>, and <em>c </em>of the parabola equation

<em> $a=4, b=-6, \:and \:c=-8$ </em>

2. You can apply this formula to find x-coordinate of the vertex

$x=-\frac{b}{2a}$ , so

$x=-\frac{-6}{2\cdot 4}\\x=\frac{3}{4}$

3. To find the y-coordinate of the vertex you use the parabola equation and x-coordinate of the vertex ( $f(-\frac{b}{2a})=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c$ )

$f(-\frac{b}{2a})=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c\\f(\frac{3}{4})=4\cdot (\frac{3}{4})^2-6\cdot (\frac{3}{4})-8\\y=\frac{-41}{4}$

To find the vertical intercept you need to evaluate x = 0 into the parabola equation

$f(x)=4x^2-6x-8\\f(0)=4(0)^2-6\cdot 0-0\\f(0)=-8$

To find the coordinates of the two intercepts of the parabola you need to solve the parabola by completing the square

$\mathrm{Add\:}8\mathrm{\:to\:both\:sides}$

$x^2-6x-8+8=0+8$

$\mathrm{Simplify}$

$4x^2-6x=8$

$\mathrm{Divide\:both\:sides\:by\:}4$

$\frac{4x^2-6x}{4}=\frac{8}{4}\\x^2-\frac{3x}{2}=2$

$\mathrm{Write\:equation\:in\:the\:form:\:\:}x^2+2ax+a^2=\left(x+a\right)^2$

$x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=2+\left(-\frac{3}{4}\right)^2\\x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{41}{16}$

$\left(x-\frac{3}{4}\right)^2=\frac{41}{16}$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

$x_1=\frac{\sqrt{41}+3}{4},\:x_2=\frac{-\sqrt{41}+3}{4}$

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Justin is asked to solve the following system of linear equations using the elimination method.

romanna [79]

Alright, lets get started.

Justin is asked to solve the linear equations using elimination method.

By using elimination method means we have to multiply some numbers in our given equations in such a way that the co-efficient of x or y become same in both equations so that we could add or subtract them to cancel one of the term either x or y.

So, given equations are :

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$-20x + 14 y = 13$

See we have 5x in first equation and -20x in second equation.

So, we try to change 5x into 20 x by multiplying it with 4, both of the equations will have 20 x in common

Lets multiply 4 in first equation

$4 * (5x-12y) = 4 * 3$

$20x - 48y = 12$

Now both equations could be added and 20 x will be cancelled out and we could easily find the value of y then solve for x.

So, Justin should try to change 5 so that it will be cancels, so option B : Answer

Hope it will help :)

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