if p(x)=ax²+bx+c and a+c=0 then the product of zeros of p(x) is equal to____
Answers
Answer:
Given:-
The external length, breadth and height of a closed rectangular wooden box are 18 cm, 10 cm and 6 cm respectively and thickness of wood is 1/2 cm. When the box is empty, it weights 15 kg and when filled with sand it weighs 100 kg
To Find:-
The weight of the cubic cm of wood and cubic cm of sand.
Solution:-
\begin{gathered} \rm \: Thickness \: of \: wood = \frac{1}{2} \: cm \\ \rm \: Internal \: length \: of \: wooden \: box = 18 - \bigg( \frac{1}{2} + \frac{1}{2} \bigg) = 18 - 1 = 17 \: cm \\ \\ \rm \: Internal \: breadth \: of \: wooden \: box = 10 - \bigg( \frac{1}{2} + \frac{1}{2} \bigg) = 9 \: cm \\ \\ \rm \: Internal \: depth \: of \: wooden \: box = 6 - \bigg( \frac{1}{2} + \frac{1}{2} \bigg) = 5 \: cm \\ \\ \therefore \: \rm Internal \: volume \: of \: wooden \: box \\ \leadsto \rm \: length \times breadth \times height \\ \leadsto \rm \: (17 \times 9 \times 5) \: {cm}^{3} \\ \\ \rm \: External \: volume \: of \: wood \\ \leadsto \rm \: length \times breadth \times height \\ \leadsto \rm \: (18 \times 10 \times 6) \: {cm}^{3} \\ \rm \leadsto \: 1080 \: {cm}^{3} \\ \\ \rm \: Volume \: of \: wood \: = \: External \: volume - Internal \: volume \\ \rm \longrightarrow \: (1080 - 765) \: {cm}^{3} = 315 \: {cm}^{3} \\ \\ \rm \: \: Weight \: of \: empty \: box = 15 \: kg \\ \implies \: \rm \: Weight \: of \: 315 \: {cm}^{3} wood \: is \: 15 \: kg \\ \therefore \: \rm Weight \: of \: 1 \: {cm}^{3} \: of \: wood = \bigg(\frac{15}{315} \bigg) \: kg = \frac{1}{21} \: kg \\ \tt \: Now \\ \rm \: Volume \: of \: sand = Internal \: volume \: of \: box = 765 \: {cm}^{3} \\ \rm Weight \: of \: sand = Weight \: of \: box \: filled \: with \: sand - Weight \: of \: empty \: box \\ \dashrightarrow \rm \: (100 - 15) \: kg = 85 \: kg \\ \\ \rm Volume \: of \: sand = 765 \: {cm}^{3} \\ \bigstar \red{ \boxed{ \purple{ \rm \: Weight \: of \: 1 \: {cm}^{3} \: of \: sand = \bigg(\frac{85}{725} \bigg) \: kg= \bold{ \frac{1}{9} < /u > < u > kg < /u > < u > < /u > < u > }}}}\end{gathered}
Thicknessofwood=
2
1
cm
Internallengthofwoodenbox=18−(
2
1
+
2
1
)=18−1=17cm
Internalbreadthofwoodenbox=10−(
2
1
+
2
1
)=9cm
Internaldepthofwoodenbox=6−(
2
1
+
2
1
)=5cm
∴Internalvolumeofwoodenbox
⇝length×breadth×height
⇝(17×9×5)cm
3
Externalvolumeofwood
⇝length×breadth×height
⇝(18×10×6)cm
3
⇝1080cm
3
Volumeofwood=Externalvolume−Internalvolume
⟶(1080−765)cm
3
=315cm
3
Weightofemptybox=15kg
⟹Weightof315cm
3
woodis15kg
∴Weightof1cm
3
ofwood=(
315
15
)kg=
21
1
kg
Now
Volumeofsand=Internalvolumeofbox=765cm
3
Weightofsand=Weightofboxfilledwithsand−Weightofemptybox
⇢(100−15)kg=85kg
Volumeofsand=765cm
3
★
Weightof1cm
3
ofsand=(
725
85
)kg=
9
1
</u><u>kg</u><u></u><u>
