70 POINTS SHOW WORK

2 answers:

8 0

Area of triangle = 1/2 x base x height

so old triangle we have 1/2 x b x h = 48

new triangle has base 3 times as long and height is half as long
so we would get:

1/2 x (3B) x (1/2H) = area

rewrite as 1/2 x 3/2 x b x H = area

divide by original are os 1/2 x b x h and you end up with 3/2

so new area = 3/2 x old area

new area = 3/2 * 48 = 72 square inches

Firdavs [7]4 years ago

6 0

Assuming that the base can be 2 in and the height can 48. changing the lengths to a base of 6 and a height of 24 the triangle's area is 72 or times the original area by the increase over the decrease ... 3/2

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Solve for x: 4 over x plus 4 over quantity x squared minus 9 equals 3 over quantity x minus 3. (2 points) Select one: a. x = -4

Kay [80]

Answer:

Step-by-step explanation:

$\dfrac{4}{x}+\dfrac{4}{x^2-9}=\dfrac{3}{x-3}$

Domain:

$x\neq0\ \wedge\ x^2-9\neq0\ \wedge\ x-3\neq0\\\\x\neq0\ \wedge\ x\neq\pm3$

solution:

$\dfrac{4}{x}+\dfrac{4}{x^2-3^2}=\dfrac{3}{x-3}$

use (a - b)(a + b) = a² - b²

$\dfrac{4}{x}+\dfrac{4}{(x-3)(x+3)}=\dfrac{3}{x-3}$

multiply both sides by (x - 3) ≠ 0

$\dfrac{4(x-3)}{x}+\dfrac{4(x-3)}{(x-3)(x+3)}=\dfrac{3(x-3)}{x-3}$

cancel (x - 3)

$\dfrac{4(x-3)}{x}+\dfrac{4}{x+3}=3$

subtract $\frac{4(x-3)}{x}$ from both sides

$\dfrac{4}{x+3}=3-\dfrac{4(x-3)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x}{x}-\dfrac{(4)(x)+(4)(-3)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x-\bigg(4x-12\bigg)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x-4x-(-12)}{x}\\\\\dfrac{4}{x+3}=\dfrac{-x+12}{x}$