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Vesnalui [34]
3 years ago
14

1234 please help please

Mathematics
2 answers:
Debora [2.8K]3 years ago
6 0
1 . 163 just subtract
ANEK [815]3 years ago
6 0
This for #2
S1-6 rows is 5 cars so that mean is 30 cars in 6 rows
S2- So his sister take and push in 3 rows, that mean still have 30 cars but no more 5 cars in a row.
S3-30 divided 3 because total is 30 cars in 6 rows nut niw is 3 rows so 30/3=10 cars/ 1 rows
I hope you understand!Good Luck!
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The volume of a cube is related to the area of s face by the formula v= a^3/2. What is the volume of a cube whose face has an ar
Sophie [7]

Answer:

  1000 m³

Step-by-step explanation:

Put the number into the formula and do the arithmetic.

  v = a^(3/2) = (100 m²)^(3/2) = (√100)³ m³ = 1000 m³

The volume is 1000 cubic meters.

4 0
3 years ago
Write two subtraction facts related to 7 +5=12
Lilit [14]

There are different ways you can write these numbers all you do is go by what the number says and think about it.

EX1) 12-5=7

EX2) 12-7=5

5 0
3 years ago
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What is the Volume of the square pyramid? (Round to the nearest tenth as needed)
goldenfox [79]

Answer:

5229 mm³

Step-by-step explanation:

Volume of square pyramid

\sf \boxed{Volume \ of \ the \ square \ pyramid = \dfrac{1}{3}*b^2*H}

      b = base length = 25 mm

       H = height

         We have to find 'H' using Pythagorean theorem,

              slant height (hypotenuse) = 28 mm

                                                  leg₁ = base length  ÷ 2 = 25÷2 = 12.5

                                                  leg₂ = H

     H² + (12.5)² = 28²

                  H²   = 784 - 156.25

                         = 627.75

                   \sf H = \sqrt{627.75}

                 H = 25.01 mm

          \sf  \text{Volume of square pyramid =$\dfrac{1}{3}*25*25*25.1$ }

                                                     = 5229 mm³

                                                     

       

\sf \boxed{\text{Volume of square pyramid = \dfrac{1}{3}*25*25*28}}

3 0
2 years ago
Read 2 more answers
Determine formula of the nth term 2, 6, 12 20 30,42​
nalin [4]

Check the forward differences of the sequence.

If \{a_n\} = \{2,6,12,20,30,42,\ldots\}, then let \{b_n\} be the sequence of first-order differences of \{a_n\}. That is, for n ≥ 1,

b_n = a_{n+1} - a_n

so that \{b_n\} = \{4, 6, 8, 10, 12, \ldots\}.

Let \{c_n\} be the sequence of differences of \{b_n\},

c_n = b_{n+1} - b_n

and we see that this is a constant sequence, \{c_n\} = \{2, 2, 2, 2, \ldots\}. In other words, \{b_n\} is an arithmetic sequence with common difference between terms of 2. That is,

2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2

and we can solve for b_n in terms of b_1=4:

b_{n+1} = b_n + 2

b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2

b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2

and so on down to

b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)

We solve for a_n in the same way.

2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)

Then

a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)

a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))

a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))

and so on down to

a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2

\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}

6 0
2 years ago
Solve the equation.<br><br> -7b=21
Kipish [7]

Answer:

b=-3

Step-by-step explanation:

-7b=21

-7b/-7=21/-7

b=-3

7 0
3 years ago
Read 2 more answers
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