Please solve Q 1. 2. 3. 4.
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1 a,2a3c this is the wright answer
Saykasayyed:
1=d
Answered by
1
1):-
let the sides of the two cubes be A and B respectively,
then according to the question, we have
6A²/6B² = 49/64,
A²/B²=49/64,
then
A/B=7/8,
therefore
ratio of volumes=A³/B³,
=7³/8³,
=343/512,
2):-
let the edge of the new cube be A,
then according to the question, we have
6³+7³+8³=A³,
216+343+512=A³,
A³=1071,
then
A=√1071,
A=10.3 approximately,
3):-
since
1 m³ = 1000 ltr
then
4 m³ = 4000 ltr,
4):-
let the radius of the circle be 10A,
then
initial area=π(10A)² = 100π A²,
new radius =9A,
then
new area = π(9A)² = 81π A²,
then
change in area = 100π A² - 81π A²,
= 19π A²,
therefore
% change = change in area/initial area × 100,
= 19π A²/100π A² × 100%,
= 19%
let the sides of the two cubes be A and B respectively,
then according to the question, we have
6A²/6B² = 49/64,
A²/B²=49/64,
then
A/B=7/8,
therefore
ratio of volumes=A³/B³,
=7³/8³,
=343/512,
2):-
let the edge of the new cube be A,
then according to the question, we have
6³+7³+8³=A³,
216+343+512=A³,
A³=1071,
then
A=√1071,
A=10.3 approximately,
3):-
since
1 m³ = 1000 ltr
then
4 m³ = 4000 ltr,
4):-
let the radius of the circle be 10A,
then
initial area=π(10A)² = 100π A²,
new radius =9A,
then
new area = π(9A)² = 81π A²,
then
change in area = 100π A² - 81π A²,
= 19π A²,
therefore
% change = change in area/initial area × 100,
= 19π A²/100π A² × 100%,
= 19%
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