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3 years ago

Find the value of x. Round the length to the nearest tenth, & plz show your work.

Mathematics

1 answer:

Helga [31]3 years ago

6 0

Answer:

x= 127.8m (nearest tenth)

Step-by-step explanation:

Please see the attached picture for the full solution.

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D. if a number has factors other than one and itself, then it is not a prime number

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2 years ago

B) Re-write the statement 16 + 5 + 2 - 1 by including one pair of brackets

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B) 16 + (5 + 2) - 1

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3 years ago

The lengths of the diagonals of a rhombus are 4x and 8x. What algebraic expression gives the perimeter of the rhombus?

Marizza181 [45]

Answer:

Approximately 17.88x or $8x\sqrt{5}$

Step-by-step explanation:

Use pythagorean formula. In a rhombus the diagonals bisect each other and they are perpendicular, so you could have a right triangle with legs of 2x and 4x, the hypoteneuse would then be $\sqrt{(2x)^{2} +(4x)^{2} }= \sqrt{4x^{2} +16x^{2}}$ which is approximately 4.47x. In a rhombus all 4 sides are the same, so multiply that by 4 and you get the perimeter. 4(4.47x) = 17.88x or if you simplify the radical instead it's $8x\sqrt{5}$

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3 years ago

Plzzz I need help with this question I tried to solve it many times but I can't

blsea [12.9K]

Answer and Step-by-step explanation: Area of a right triangle, (as any other triangle), is calculated as: $A1=\frac{(base)(height)}{2}$

Area of a rectangle is calculated as: $A2=(side)(side)$

Area of a right trapezoid is: $A3=\frac{(a+b)h}{2}$ , where:

a is short base

b is long base

h is height

1) Expressing areas in terms of x:

Area of triangle S1:

$S1=\frac{(2x-3)(4x-6)}{2}$

$S1=4x^{2}-12x+9$

Area of rectangle S2:

$S2 = (4x-6)(3x-2)$

$S2=12x^{2}-26x+12$

Area of trapezoid S3:

$S3=\frac{(2x+3+4x+1)(2x-3)}{2}$

$S3=\frac{(6x+4)(2x-3)}{2}$

$S3=6x^{2}-5x-6$

2) a) $S=4x^{2}-12x+9+12x^{2}-26x+12-(6x^{2}-5x-6)$

$S=4x^{2}-12x+9+12x^{2}-26x+12-6x^{2}+5x+6$

$S=10x^{2}-33x+37$

Which is the same as S = (2x-3)(5x-9)

b) For the areas to be the same:

$\frac{(3x-2+3x-2+2x-3)(4x-6)}{2}=\frac{(6x+4)(2x-3)}{2}$

$\frac{(8x-7)(4x-6)}{2}=\frac{(6x+4)(2x-3)}{2}$

$32x^{2}-48x-28x+42=12x^{2}+8x-18x-12$

$20x^{2}-66x+54=0$

Using Bhaskara to solve the second degree equation:

$\frac{66+\sqrt{(-66)^{2}-(4.20.54)} }{2(20)}$

$x_{1}=\frac{66+6}{40}$ = 1.8

$x_{2}=\frac{66-6}{40}$ = 1.5

For the areas of AFGC and ADEB to be equal, x has to be 1.5 or 1.8.

c) Expand a polynomial (or equation) is to multiply all the terms, remiving the parenthesis. Reduce a polynomial (or equation) is to combine terms alike,e.g.:

$S=(2x-3)(5x-9)$