C-11-Q043mediumsingle_mcq

The proof of invertendo from $\dfrac{a}{b} = \dfrac{c}{d}$ uses the step—

amultiply both sides by $bd$ to get $ad = bc$ , then divide by $ac$
badd $1$ to both sides

ব্যাখ্যা

Starting from $\dfrac{a}{b}=\dfrac{c}{d}$ , cross-multiplying (multiplying by ) gives . Dividing this through by isolates the terms as , which is exactly the invertendo relation.

এই অধ্যায়ের আরও প্রশ্ন

C-11-Q044The proof of alternendo uses the step—
C-11-Q045If , by alternendo we get—
C-11-Q046If , by invertendo we get—
C-11-Q047From , by combining alternendo and invertendo we obtain—
C-11-Q048If , then by alternendo
C-11-Q049Componendo: if , then—
C-11-Q050Dividendo: if , then—
C-11-Q051Componendo-Dividendo: if , then—
C-11-Q052Componendo is proved by—

The proof of invertendo from $\dfrac{a}{b} = \dfrac{c}{d}$ uses the step—

amultiply both sides by

bd

to get

ad = bc

, then divide by

ac

badd

1

to both sides

ব্যাখ্যা

Starting from $\dfrac{a}{b}=\dfrac{c}{d}$ , cross-multiplying (multiplying by ) gives . Dividing this through by isolates the terms as , which is exactly the invertendo relation.

The proof of invertendo from ab=cd\dfrac{a}{b} = \dfrac{c}{d}ba​=dc​ uses the step—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The proof of invertendo from ab=cd\dfrac{a}{b} = \dfrac{c}{d}ba​=dc​ uses the step—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The proof of invertendo from $\dfrac{a}{b} = \dfrac{c}{d}$ uses the step—

The proof of invertendo from $\dfrac{a}{b} = \dfrac{c}{d}$ uses the step—