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katrin2010 [14]
3 years ago
15

What number has the same value as12 tens

Mathematics
1 answer:
Nuetrik [128]3 years ago
7 0
12 tens as in 12 10's
that equals
10+10+10+10+10+10+10+10+10+10+10+10
or 12 times 10
when multiply by 10, just move decimal place on e to the right
12.0 times 10=120

the answe ris 120
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Santa has lots of mixed up socks in his sack! If he has 6 green socks, 4 gold socks, 8 black socks and 2 red socks, what is the
olganol [36]
Add it all together and divide it by 2 the answer is 10

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4 years ago
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On a map, Mary's house is located at (−3, 1) and the grocery store is located at (5, 7). Mary walks on a straight line from her
FrozenT [24]

The ordered pair that  represents the halfway point between Mary's home and the grocery store is (1, 4)

The midpoint (a, b) between two linear points (x_1, y_1) and (x_2, y_2) is calculated as:

a=\frac{x_1+x_2}{2} \\\\b=\frac{y_1+y_2}{2}

On a map, Mary's house is located at (−3, 1) and the grocery store is located at (5, 7).

x_1=-3, y_1=1, x_2=5, y_2=7

Substitute these values into the mid-point formulae

a=\frac{-3+5}{2} \\\\a=\frac{2}{2}\\\\a=1

b=\frac{1+7}{2} \\\\b=\frac{8}{2}\\\\b=4

The ordered pair that  represents the halfway point between Mary's home and the grocery store is (1, 4)

Learn more on the midpoint of a line here: brainly.com/question/18315903

5 0
2 years ago
Alvin's age is two times Elga's age. The sum of their ages is 24. What is Elga's age?
gulaghasi [49]
Alvin is 16 and Elga is 8.

16 is two times bigger than 8 and 16+8= 24
8 0
2 years ago
Read 2 more answers
Use induction to show that 12 + 22 + 32 + ... + n2 = n(n+1)(2n+1)/6, for all n > 1.
dlinn [17]

Answer with Step-by-step explanation:

Let P(n)=1^2+2^2+3^2+.....+n^2=\frac{n(n+1)(2n+1)}{6}

Substitute n=2

Then  P(2)=1+2^2=5

P(2)=\frac{2(2+1)(4+1)}{6}=5

Hence, P(n) is true for n=2

Suppose that P(n) is true for n=k >1

P(k)=1^2+2^2+3^2+...+k^2=\frac{k(k+1)(2k+1)}{6}

Now, we shall prove that p(n) is true for n=k+1

P(k+1)=1^2+2^2+3^2+...+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}

LHS

P(k+1)=1^2+2^2+3^2+.....+k^2+(k+1)^2

Substitute the value of P(k)

P(k+)=\frac{k(k+1)(2k+1)}{6}+(k+1)^2

P(k+1)=(k+1)(\frac{k(2k+1}{6})+k+1)

P(k+1)=(k+1)(\frac{2k^2+k+6k+6}{6})

P(k+1)=(k+1)(\frac{2k^2+7k+6}{6})

P(k+1)=(k+1)(\frac{2k^2+4k+3k+6}{6})

P(k+1)=\frac{(k+1)(k+2)(2k+3)}{6}

LHS=RHS

Hence, P(n) is true for all n >1.

Hence, proved

4 0
3 years ago
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