What is the value of x?
х= 12
х= 14
x= 22
х= 24

Answers

i think the correct answer is x:14

Explanation:

hahl

Step-by-step explanation:

x=22

Step-by-step explanation:

According to the Alternate Interior angles theorem, the bottom left angle and the upper right angle have the same measure. Because they were formed from a pair of parallel lines crossed by a transversal one.

So, algebraically:

5x+5=115\\ 5x+5-5=115-5\\ 5x=110\\ \frac{5x}{5}=\frac{115}{5}\\x=22


Two parallel lines are crossed by a transversal. horizontal and parallel lines b and c are cut by tr

x=22

Step-by-step explanation:

According to the Alternate Interior angles theorem, the bottom left angle and the upper right angle have the same measure. Because they were formed from a pair of parallel lines crossed by a transversal one.

So, algebraically:

5x+5=115\\ 5x+5-5=115-5\\ 5x=110\\ \frac{5x}{5}=\frac{115}{5}\\x=22


Two parallel lines are crossed by a transversal. horizontal and parallel lines b and c are cut by tr

383.08‬

Step-by-step explanation:

Multiply 3 times 122 which is 366

Then multiply .14 times 122 which is 17.08

Then add.

308

Step-by-step explanation:

i looked it up

22

Step-by-step explanation:

12

Step-by-step explanation:

The mass of Ni(CO)_4 allowable in the laboratory is 4599.5 grams

Explanation:

To calculate the volume of cuboid, we use the equation:

V=l\times b\times h

where,

V = volume of cuboid

l = length of cuboid = 14 ft

b = breadth of cuboid = 22 ft

h = height of cuboid = 9 ft

Putting values in above equation, we get:

V=14\times 22\times 9=2772ft^3=78503.04L     (Conversion factor:  1ft^3=28.32L

To calculate the moles of gas, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 1.00 atm

V = Volume of the gas = 78503.04 L

T = Temperature of the gas = 23^oC=[23+273]K=296K

R = Gas constant = 0.0821\text{ L. atm }mol^{-1}K^{-1}

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

1.00atm\times 78503.04L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 296K\\\\n=\frac{1.00\times 78503.04}{0.0821\times 296}=3230.4mol

Applying unitary method:

For every 109 moles of gas, the moles of Ni(CO)_4 present are 1 moles

So, for 3230.4 moles of gas, the moles of Ni(CO)_4 present will be = \frac{1}{109}\times 3230.4=26.94mol

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of Ni(CO)_4 = 26.94 moles

Molar mass of Ni(CO)_4 = 170.73 g/mol

Putting values in above equation, we get:

26.94mol=\frac{\text{Mass of }Ni(CO)_4}{170.73g/mol}\\\\\text{Mass of }Ni(CO)_4=(26.94mol\times 170.73g/mol)=4599.5g

Hence, the mass of Ni(CO)_4 allowable in the laboratory is 4599.5 grams

answer:

hxjjxdxuxnkc ckcnkckc ckc

step-by-step explanation:

shjdicmckcmchzjxixbxj c



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