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maxonik [38]
3 years ago
14

An elementary school enrolled 57 students in each grade from kindergarten through sixth. How many students are there in the scho

ol
Mathematics
2 answers:
Luba_88 [7]3 years ago
6 0

Step-by-step explanation:

From kindergarten to sixth grade is seven grades and if for each grade 57 new students get enrolled then 57*7=399 there are 399 students

Tresset [83]3 years ago
5 0

Answer:

399

Step-by-step explanation:

57x7=399

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What is the distance between P(12, 4) and Q(-8, 2)
abruzzese [7]

\huge{\boxed{2 \sqrt{101}}}

The distance formula is \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}, where (x_1, y_1) and (x_2, y_2) are the points.

Substitute in the points. \sqrt{(2-4)^2+(-8-12)^2}

Subtract. \sqrt{(-2)^2+(-20)^2}

Solve the exponents. \sqrt{4+400}

Add. \sqrt{404}

Now, we can simplify this square root just a little bit. 404 has a square factor of 4. \sqrt{404}=\sqrt{4}*\sqrt{101}

The square root of 4 equals 2. \boxed{2 \sqrt{101}}

6 0
3 years ago
Read 2 more answers
1+1=21+1=21+1=21+1=21+1=21+1=21+1=21+1=21+1=21+1=21+1=2<br><br><br> lol
Vilka [71]

Answer:

1 + 1 equals 2.

Step-by-step explanation:

When you take one of something, and you add it to another 1 of something, then you get 2 of something.

7 0
3 years ago
Read 2 more answers
The answer to a division problem in David's homework was 12. Which of the following expressions might he have solved? Select ALL
emmainna [20.7K]
None of those equal 12?
4 0
2 years ago
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What is the greatest integer a, such that a^2 + 3b is less than (2b)^2, assuming that b is 5? Please answer and have a nice day!
solniwko [45]
<h3>Answer: Largest value is a = 9</h3>

===================================================

Work Shown:

b = 5

(2b)^2 = (2*5)^2 = 100

So we want the expression a^2+3b to be less than (2b)^2 = 100

We need to solve a^2 + 3b < 100 which turns into

a^2 + 3b < 100

a^2 + 3(5) < 100

a^2 + 15 < 100

after substituting in b = 5.

------------------

Let's isolate 'a'

a^2 + 15 < 100

a^2 < 100-15

a^2 < 85

a < sqrt(85)

a < 9.2195

'a' is an integer, so we round down to the nearest whole number to get a \le 9

So the greatest integer possible for 'a' is a = 9.

------------------

Check:

plug in a = 9 and b = 5

a^2 + 3b < 100

9^2 + 3(5) < 100

81 + 15 < 100

96 < 100 .... true statement

now try a = 10 and b = 5

a^2 + 3b < 100

10^2 + 3(5) < 100

100 + 15 < 100 ... you can probably already see the issue

115 < 100 ... this is false, so a = 10 doesn't work

5 0
3 years ago
5,286÷3 how many digits will the classroom number have
docker41 [41]
<span>5,286÷3 how many digits will the classroom number have
Notice that we have 4 digits as dividend with a place value of thousands as the highest and to be divided with our divisor that have only 1 digit with a place value of ones.
Now, let’s see how many digit will our quotient have:
=> 5 286 / 3
=> 1 762 is the quotient, it has still 4 digit with a place value of thousands.
To check simply multiply our quotient and divisor.</span>



5 0
3 years ago
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