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3 years ago

The area of a circle is 497 cm2. What is the circumference, in centimeters? Express

Mathematics

1 answer:

SVEN [57.7K]3 years ago

4 0

Answer:

14π

Step-by-step explanation:

Area πr^2= 49π sq cm

so r= 7 cm

perimeter= 2πr= 14π

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Find the measure of each angle in the triangle.

ValentinkaMS [17]

Answer:

x=26°

Step-by-step explanation:

3x-14+x+90=180

4x-14=90

4x=104

x=104/4

x=26°

6 0

3 years ago

Which of the following is equivalent to 5 · (7 · 4)? *

MariettaO [177]

B because you multiply 7 times 4 that’s 28 then times 5 that’s 140 so for b you do 5 times 7 that’s 35 then times 4 that’s 140 so b is the correct answer, hope it helps

8 0

3 years ago

I WILL MARK THE CORRECT ALPHA!!

Kitty [74]

Answer:

I think the answer is Step 2, 12x − 16 = 16 + 4x

plz mark me as brainliest if i was right :)

4 0

3 years ago

I need help.. i really want to go sleep.. thank you so much...

Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given $\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$

To verify that the above equality is true or false:

Now find $\sum\limits_{k=0}^8\frac{1}{k+3}$

Expanding the summation we get

$\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3}$ $\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Now find $\sum\limits_{i=3}^{11}\frac{1}{i}$

Expanding the summation we get

$\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Comparing the two series we get,

$\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$ so the given equality is true.

2) Given $\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}$

Verify the above equality is true or false

Now find $\sum\limits_{k=0}^4\frac{3k+3}{k+6}$

Expanding the summation we get

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}$

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}$

now find $\sum\limits_{i=1}^3\frac{3i}{i+5}$

Expanding the summation we get

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}$

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}$

Comparing the series we get that the given equality is false.

ie, $\sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}$

6 0

3 years ago

Which expression is not equivalent to 8

Tom [10]

12 + n

if you substitute n for any whole number and simplify. you will find that 12+n does not equal any others. for example, i used 6 in place of n and all equations gives you 10 except 12+n

hope this helps!

5 0

2 years ago