Answer:
x = 44/19, y = 33/20
Step-by-step explanation:
Multiply the first eq by 18, to get:
90/y - 36/x = 39
so 36/x = 90/y - 39
in this form we can plug it into the second eq:
90/y - 39 - 24/y = 1 =>
66/y = 40 =>
y = 66/40 = 33/20
Fill that in the first:
100/33 - 2/x = 13/6 =>
200/66 - 143/66 = 2/x =>
57/66 = 2/x =>
57x = 2*66
x = 132/57 = 44/19
<em>ANSWER CORRECTLY ASAP </em>
<em>ANSWER CORRECTLY ASAP 1. An automobile tire has a pressure of 210.0 kPa at 20.0oC. What will be the tire pressure after diving, if the tire temperature rises to 35.0oC? (221 kPa)</em>
<em>ANSWER CORRECTLY ASAP 1. An automobile tire has a pressure of 210.0 kPa at 20.0oC. What will be the tire pressure after diving, if the tire temperature rises to 35.0oC? (221 kPa)2. If a sample of gas has a volume of 355 mL at 44.1oC and 87.2 kPa, what will its volume be at STP? (263 mL)</em>
<em>ANSWER CORRECTLY ASAP 1. An automobile tire has a pressure of 210.0 kPa at 20.0oC. What will be the tire pressure after diving, if the tire temperature rises to 35.0oC? (221 kPa)2. If a sample of gas has a volume of 355 mL at 44.1oC and 87.2 kPa, what will its volume be at STP? (263 mL)3. A tank of compressed gas has a maximum safe pressure limit of 826 kPa. The pressure gauge reads 388 kPa when the temperature is 24.1oC. What is the highest temperature the tank can withstand safely? (632 K)</em>
<em>ANSWER CORRECTLY ASAP 1. An automobile tire has a pressure of 210.0 kPa at 20.0oC. What will be the tire pressure after diving, if the tire temperature rises to 35.0oC? (221 kPa)2. If a sample of gas has a volume of 355 mL at 44.1oC and 87.2 kPa, what will its volume be at STP? (263 mL)3. A tank of compressed gas has a maximum safe pressure limit of 826 kPa. The pressure gauge reads 388 kPa when the temperature is 24.1oC. What is the highest temperature the tank can withstand safely? (632 K)4. A gas occupies a volume of 560 cm3 at a temperature of 120.0oC. To what temperature must the gas be lowered, if it is to occupy 400.0 cm3? Assume a constant pressure.</em>
<em>ANSWER CORRECTLY ASAP 1. An automobile tire has a pressure of 210.0 kPa at 20.0oC. What will be the tire pressure after diving, if the tire temperature rises to 35.0oC? (221 kPa)2. If a sample of gas has a volume of 355 mL at 44.1oC and 87.2 kPa, what will its volume be at STP? (263 mL)3. A tank of compressed gas has a maximum safe pressure limit of 826 kPa. The pressure gauge reads 388 kPa when the temperature is 24.1oC. What is the highest temperature the tank can withstand safely? (632 K)4. A gas occupies a volume of 560 cm3 at a temperature of 120.0oC. To what temperature must the gas be lowered, if it is to occupy 400.0 cm3? Assume a constant pressure.5. If 122 mL of oxygen are collected at 27.5oC and 740 mmHg, what will the volume be of the dry gas at STP? (760 mmHg = 101.3 kPa)</em>
<em>ANSWER CORRECTLY ASAP 1. An automobile tire has a pressure of 210.0 kPa at 20.0oC. What will be the tire pressure after diving, if the tire temperature rises to 35.0oC? (221 kPa)2. If a sample of gas has a volume of 355 mL at 44.1oC and 87.2 kPa, what will its volume be at STP? (263 mL)3. A tank of compressed gas has a maximum safe pressure limit of 826 kPa. The pressure gauge reads 388 kPa when the temperature is 24.1oC. What is the highest temperature the tank can withstand safely? (632 K)4. A gas occupies a volume of 560 cm3 at a temperature of 120.0oC. To what temperature must the gas be lowered, if it is to occupy 400.0 cm3? Assume a constant pressure.5. If 122 mL of oxygen are collected at 27.5oC and 740 mmHg, what will the volume be of the dry gas at STP? (760 mmHg = 101.3 kPa)6. A sample of neon gas has a volume of 0.261 L and a pressure of 108.5 kPa. What volume will the gas occupy at 96.5 kPa, if the temperature remains constant?</em>
Answer:
3
Step-by-step explanation:
Answer:
Yes
Step-by-step explanation:
60x means 60*x.
When there is nothing between a number and variable it means multiplication.
Answer:
2 bottles
Step-by-step explanation:
The 15 trees will need ...
(15 trees)×(10.5 fl oz/tree) = 157.5 fl oz
This is about ...
(157.5 fl oz)/(33.814 fl oz/L) ≈ 4.66 L
This is just slightly more than the quantity available in 1 bottle. It amounts to ...
4.66 L/(4.1 L/bottle) ≈ 1.14 bottles
We assume that only whole bottles can be purchased, and that we do not have an inventory of fertilizer on hand.
2 bottles are needed to fertilize 15 trees
