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2 years ago

What is the length of the hypotenuse of ABC

Mathematics

1 answer:

tigry1 [53]2 years ago

4 0

Answer:

c＝√(a²+b²)

C is hypotenuse

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PLEASE HELP WILL GIVE BRAINLIEST AND 5.0 RATING

denis23 [38]

Answer:

Step-by-step explanation:

v is 5/-2 on the graph

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3 years ago

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Are the following lines parallel perpendicular or neither y=-5/3x - 5 and y=4/3x - 5

katrin [286]

Answer:

neither

Step-by-step explanation:

For lines to be parallels the slopes must be equal, for the lines to be perpendicular the lines have to be inverse reciprocals meaning if you have a slope of 1/2X then the perpendicular line must be -2X

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2 years ago

Help guys pls 22x²+13x+1

myrzilka [38]

Answer:

(11x+1)(2x+1)

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3 years ago

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BabaBlast [244]

$\underline{\bf{Given \:equation:-}}$

$\\ \sf{:}\dashrightarrow ax^2+by+c=0$

$\sf Let\:roots\;of\:the\: equation\:be\:\alpha\:and\beta.$

$\sf We\:know,$

$\boxed{\sf sum\:of\:roots=\alpha+\beta=\dfrac{-b}{a}}$

$\boxed{\sf Product\:of\:roots=\alpha\beta=\dfrac{c}{a}}$

$\underline{\large{\bf Identities\:used:-}}$

$\boxed{\sf (a+b)^2=a^2+2ab+b^2}$

$\boxed{\sf (√a)^2=a}$

$\boxed{\sf \sqrt{a}\sqrt{b}=\sqrt{ab}}$

$\boxed{\sf \sqrt{\sqrt{a}}=a}$

$\underline{\bf Final\: Solution:-}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}$

$\bull\sf Apply\: Squares$

$\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2= (\sqrt{\alpha})^2+2\sqrt{\alpha}\sqrt{\beta}+(\sqrt{\beta})^2$

$\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2 \alpha+\beta+2\sqrt{\alpha\beta}$

$\bull\sf Put\:values$

$\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2=\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\sqrt{\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}}$

$\bull\sf Simplify$

$\\ \sf{:}\dashrightarrow \underline{\boxed{\bf {\sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\sqrt{\dfrac{-b}{a}}+\sqrt{2}\dfrac{c}{a}}}}$

$\underline{\bf More\: simplification:-}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{-b}}{\sqrt{a}}+\dfrac{c\sqrt{2}}{a}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{a}\sqrt{-b}+c\sqrt{2}}{a}$

$\underline{\Large{\bf Simplified\: Answer:-}}$

$\\ \sf{:}\dashrightarrow\underline{\boxed{\bf{ \sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\dfrac{\sqrt{-ab}+c\sqrt{2}}{a}}}}$

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2 years ago

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How is 3/5 and 8/10 Are equivalent fractions?

olya-2409 [2.1K]

Hi!

These are not equivalent fractions because 8/10 reduces to 4/5, not 3/5, therefore, they are not equivalent.

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