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otez555 [7]
2 years ago
5

A cone has a radius of 10 and a height of 3. What is the volume of the cone?

Mathematics
1 answer:
MissTica2 years ago
5 0

Hope this helps you

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Sheila eats 3/4 of a bag of carrots each week, how many bags of baby carrots does she eat in 6 weeks? Write in simplest form
Alla [95]
3/4 x 6 = 18/4 = 4 1/2 

she eats 4 1/2 bags of carrots in 6 weeks
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3 years ago
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Given ABD =≈ CBD
vredina [299]

Answer:

AD BC only

Step-by-step explanation:

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i need help on this question it says yunna missed 5 points out of 100 points on her math test.What decimal number repesents the
kap26 [50]
If she got 5/100 wrong she got 95%
95% is the same as 0.95
6 0
3 years ago
X is the sum of y consecutive integers. w is the sum of z consecutive integers. If y = 2z, and y and z are both positive integer
zlopas [31]

Answer:

The answer is C.

Step-by-step explanation:

The key to solve the problem is the equation given

y = 2z         which means that<u> y is always an even number.</u>

For example:

let z = 1;       y  = 2(1)     y = 2;

let z = 2;      y  = 2(2)     y = 4;

let z = 3;      y  = 2(3)     y = 6;

And we know that for any set of consecutive integers with an even number of terms, the sum of the integers is never a multiple of the number of terms.

For example:

Set of numbers: [1,2,3,4,5,6] ;

Sum of the set:  1+2+3+4+5+6 = 21 ;

Number of terms of the set: 6;

Therefore, 21 is not a multiple of 6.

But for any set of consecutive integers with an odd number of terms, the sum of the integers is always a multiple of the number of terms.

For example:

Set of numbers: [1,2,3,4,5,] ;

Sum of the set:  1+2+3+4+5 = 15 ;

Number of terms of the set: 5;

Therefore, 15 is a multiple of 5.

And that's why we know that if y is always an even number (y = 2z), the sum of the consecutive integers (x) is not a multiple of y, therefore x/y cannot be an integer.

8 0
3 years ago
Evaluate the series
dalvyx [7]

Answer:

the value of the series;

\sum_{k=1}^{6}(25-k^2) = 59

C) 59

Step-by-step explanation:

Recall that;

\sum_{1}^{n}a_n = a_1+a_2+...+a_n\\

Therefore, we can evaluate the series;

\sum_{k=1}^{6}(25-k^2)

by summing the values of the series within that interval.

the values of the series are evaluated by substituting the corresponding values of k into the equation.

\sum_{k=1}^{6}(25-k^2) =(25-1^2)+(25-2^2)+(25-3^2)+(25-4^2)+(25-5^2)+(25-6^2)\\\sum_{k=1}^{6}(25-k^2) =(25-1)+(25-4)+(25-9)+(25-16)+(25-25)+(25-36)\\\sum_{k=1}^{6}(25-k^2) =24+21+16+9+0+(-11)\\\sum_{k=1}^{6}(25-k^2) = 59\\

So, the value of the series;

\sum_{k=1}^{6}(25-k^2) = 59

6 0
3 years ago
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