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Kitty [74]
3 years ago
5

Find the volume of a ball with a radius of 7cm.

Mathematics
1 answer:
azamat3 years ago
7 0

Answer:

1436.7550402417cm³

Explanation:

Formula for volume: \bold{\frac{4}{3} \ \pi \ r^3}

π = 3.14

R = 7

4/3 · 3.14 · 343

= 1436.7550402417cm³

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Answer please hurry i need it
Snezhnost [94]

Answer:

0

Step-by-step explanation:

x+12 = 9+2x +3

x = 12-12

x = 0

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3 years ago
If I have more 15 Pokemon cards than Austin, which expression shows how many |<br>have?​
Nataly_w [17]

Answer:

a+15=x

Step-by-step explanation:

Austin is represented as variable a, and you have 15 more cards.

x is your amount of cards

5 0
2 years ago
Which expression is equivalent to 4/9x7?
LiRa [457]

Is there an answer key?

8 0
3 years ago
How many complex zeros does the polynomial function have?<br> f(x)=−3x^6−2x^4+5x+6
REY [17]

one way would be to factor

I can't factor it so we will have to use Descartes' Rule of Signs which is helpful for finding how many real roots you have


it goes like this:

for a polynomial with real coefients, consider f(x)=-3x^6-2x^4+5x+6.

after arranging the terms in decending order in terms of degree, count how many times the signs of the coeffients change direction and minus 2 from that number until you get to 1 or 0. that will be the number of even roots the function can have

We have (-, -, +, +). the signs changed 1 times, so it has 1 real positive root


to get the negative roots, we evaluate f(-x) and see how many times the root changes

f(-x)=-3x^6-2x^4-5x+6

signs are (-, -, -, +). there was 1 change in sign

so the function has 1 real negative root



a total of 2 real roots


a function of degree n can have at most, n roots


our function is degree 6 so it has 6 roots

if 2 are real, then the others must be complex

6-2=4 so there are 4 complex roots


you can also show that there are only 2 real roots by using a graphing utility to see that there are only 2 real roots

7 0
3 years ago
3)
Wewaii [24]

Answer:

For covering  1 unit area of the entire playground, the amount of sand required is equal to volume of 3 buckets of sand.

Step-by-step explanation:

Given -

\frac{1}{3} volume of sand in bucket is able to cover \frac{1}{9} area of the entire playground

Thus,

For covering  \frac{1}{9} unit area of the entire playground, the amount of sand required is equal to  \frac{1}{3}  of the total volume of sand in bucket

For covering  1 unit area of the entire playground, the amount of sand required is equal to

\frac{\frac{1}{3} }{\frac{1}{9} } \\\\\frac{1}{3}  * \frac{9}{1} \\\frac{9}{3}\\= 3

For covering  1 unit area of the entire playground, the amount of sand required is equal to volume of 3 buckets of sand.

4 0
3 years ago
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