Mathematics, 03.08.2019 20:10, jpsaad00

The number of students at nicks school has 98% of last years number. currently, there are 1,170 students. how many students were there last year? round to nearest while number.

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answered: maddietomlinson113

Let, the number of students at Nick's school on last year be x.

Given, 98% of x is the number of students in this year.

Also given, there are 1170 students in the school this year.

First we will find 98% of x.

98% of x = $(x)(\frac{98}{100})$ = $\frac{98x}{100}$

So, we can write the equation as,

$\frac{98x}{100} = 1170$

To solve it for x, first we will move 100 to the other side by multiplying 100 to both sides, we will get,

$(\frac{98x}{100}) (100)= (1170)(100)$

98x = (1170)(100)

98x = 117000

To get x, we will move 98 to the right side by dividing it to both sides. We will get,

$\frac{98x}{98} =\frac{117000}{98}$

$x =\frac{117000}{98}$

x = 1194 (Approximately taken to the nearest whole number)

We have got the required answer.

The number of students were there in the school last year is 1,194.

answered: jackparo9640

1170/98*100=1193.87 = 1194 is your Answer.

answered: Guest

9.55 degrees.

step-by-step explanation:

the given angle is $\frac{31\pi}{6}=5.166667\pi$

1 cycle is equal to 2π radians. so 4π will be equal to two cycles.

5.166667π-4π = 1.166667π

hence, this angle goes a little more than negative x axis.

1.166667π-π= 0.166667π

hence, the reference angle is 0.166667π radians.

1 radian = 57.2958 degrees

now, in degrees, this should be equals to 9.55 degrees.

answered: Guest

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.