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ivolga24 [154]
2 years ago
11

What is the best estimate length of a football

Mathematics
1 answer:
Ne4ueva [31]2 years ago
5 0
11 to 11.2 inches Is an estimate length of a football
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Explain how to solve the inequality below. (solve and explain the steps please, do not just put the answer)
Mazyrski [523]

Answer:

x> -5

Step-by-step explanation:

subtract 4 from both sides: -2x<10

multiply both sides by -1 (reverse the inequality)

2x>-10

divide both sides by two: x>-5

7 0
1 year ago
Read 2 more answers
Is the following relation a function. <br><br> A) yes<br> B) no
dmitriy555 [2]

Answer:

yes

Step-by-step explanation:

A relation is a function in which for each input there is only one output.

In a relation, y is a function of x.

Now we look at the mapping

For input 6, -1 is the output

for input -1, 2 is the output

for input 4, 3 is the output

for input 0, 3 is the output

For each input , there is an output. Input is not repeating. Input is occurring only once.

So this relation is a function.

4 0
1 year ago
What's the formula for R in C= 2•3.14r
natulia [17]
C=2\cdot3.14r\\\\C=6.28r\\\\6.28r=C\ \ \ \ |divide\ both\ sides\ by\ 6.28\\\\\huge\boxed{r=\dfrac{C}{6.28}}


C=2\pi r\\\\2\pi r=C\ \ \ \ |divide\ both\ sides\ by\ 2\pi\\\\\huge\boxed{r=\frac{C}{2\pi}}
5 0
1 year ago
Please Help !!!
solong [7]
Best Answer: 2 LiCl = 2 Li + Cl2
mass Li = 56.8 mL x 0.534 g/mL=30.3 g
moles Li = 30.3 g / 6.941 g/mol=4.37
the ratio between Li and LiCl is 2 : 2 ( or 1 : 1)

moles LiCl required = 4.37
mass LiCl = 4.37 mol x 42.394 g/mol=185.3 g


Cu + 2 AgNO3 = Cu/NO3)2 + 2 Ag
the ratio between Cu and AgNO3 is 1 : 2
moles AgNO3 required = 4.2 x 2 = 8.4 : but we have only 6.3 moles of AgNO3 so AgNO3 is the limiting reactant
moles Cu reacted = 6.3 / 2 = 3.15
moles Cu in excess = 4.2 - 3.15 =1.05

N2 + 3 H2 = 2 NH3
moles N2 = 42.5 g / 28.0134 g/mol=1.52
the ratio between N2 and H2 is 1 : 3
moles H2 required = 1.52 x 3 =4.56
actual moles H2 = 10.1 g / 2.016 g/mol= 5.00 so H2 is in excess and N2 is the limiting reactant
moles NH3 = 1.52 x 2 = 3.04
mass NH3 = 3.04 x 17.0337 g/mol=51.8 g
moles H2 in excess = 5.00 - 4.56 =0.44
mass H2 in excess = 0.44 mol x 2.016 g/mol=0.887 g
5 0
2 years ago
Three hundred four million, eight hundred thousand, four hundred in expanded form?
aleksandr82 [10.1K]
304,000,000+800,000+400 that's expanded form
8 0
1 year ago
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