Which expression will help you find the area of the triangular base?

Mathematics

1 answer:

krok68 [10]3 years ago

6 0

Hopes this helps please correct me if I am wrong:

Answer: D. 8•12

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Vinil7 [7]

Answer:

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Step-by-step explanation:

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Vera_Pavlovna [14]

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Solve for y 16y^2-25=0

Pepsi [2]

Answer:

$\large\boxed{x=-\dfrac{5}{4}\ \vee\ x=\dfrac{5}{4}}$

Step-by-step explanation:

$16y^2-25=0\\\\METHOD\ 1:\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\16=4^2\ \text{and}\ 25=5^2\ \text{therefore we have}\\\\4^2y^2-5^2=0\\\\(4y)^2-5^2=0\\\\(4y-5)(4y+5)+0\iff4y-5=0\ \vee\ 4y+5=0\\\\4y-5=0\qquad\text{add 5 to both sides}\\4y=5\qquad\text{divide both sides by 4}\\\boxed{y=\dfrac{5}{4}}\\\\4y+5=0\qquad\text{subtract 5 from both sides}\\4y=-5\qquad\text{divide both sides by 4}\\\boxed{x=-\dfrac{5}{4}}$

$METHOD\ 2:\\\\16y^2-25=0\qquad\text{add 25 to both sides}\\\\16y^2=25\qquad\text{divide both sides by 16}\\\\y^2=\dfrac{25}{16}\to y=\pm\sqrt{\dfrac{25}{26}}\\\\y=-\dfrac{\sqrt{25}}{\sqrt{16}}\ \vee\ x=\dfrac{\sqrt{25}}{\sqrt{16}}\\\\\boxed{y=-\dfrac{5}{4}\ \vee\ x=\dfrac{5}{4}}$