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Svetach [21]
3 years ago
9

What is 17,384,069 rounded to the nearest ten

Mathematics
2 answers:
lbvjy [14]3 years ago
6 0
The answer to "what is 17,384,069 rounded to the nearest ten is 17,384,070"
frosja888 [35]3 years ago
5 0
I believe it is 17,400,100. :)
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A circle has a radius that is 4 centimeters long. If a central angle has a measure of 3 radians, what is the length of the arc t
kirill115 [55]
The total length of the circle is calculated by the following formula:

Lenght = 2.π.r

As r = 4c

Then,

Lenght = 2.π.4

Lenght = 8π centimetrs

Now use the rule of 3


But we have to know:

...<..> ...

Then,

8π ________ 2π <= 360°

x ________ 3π


2π . x = 8π . 3π

x = (8π . 3π) / 2π

x = (8 . 3π)/ 2

x = 12π centimenters

x ~ 37,69 c
4 0
2 years ago
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Rob has -180 points in a game. He gains one-fourth of his points back on his next turn and gains back an additional one-fifth of
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Answer:not sure sorry

Step-by-step explanation: I need points

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2 years ago
A mixture of concrete is made up of sand and cement in the ratio of 5:3.How many cubic centimeters of sand is needed to make 160
bazaltina [42]

Answer: 100\ cm^3

Step-by-step explanation:

Given

The ratio of mixture of sand and cement is 5:3

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3 0
2 years ago
How do I solve inscribed angles?
Slav-nsk [51]
The triangles are similar so the angles are equal
2x^2 - 2 = 10x - 2
2x^2 - 2 - 10x + 2 = 0
2x^2 - 10x = 0
2x(x - 5) = 0
2x = 0       OR        x - 5 = 0
x = 0         OR        x = 5
0 doesn't make sense, because IF I plug 0 into 2x^2 - 2 I get a negative number.  We need to use 5 to find the angle measure

2(5^2) - 2 = 2(25) - 2 = 50 - 2 = 48 degrees or the measure of CED
4 0
2 years ago
Please answer correctly !!!!!!! Will mark brainliest !!!!!!!!!!!!
LUCKY_DIMON [66]

Answer:

=8\sqrt{15}b^{\frac{7}{2}}

Step-by-step explanation:

\sqrt{24b^3}\sqrt{40b^2}\sqrt{b^2}

=\sqrt{40}\sqrt{b^2}\sqrt{b^2}\sqrt{24b^3}

=\sqrt{40}b^2\sqrt{24b^3}

\sqrt{24b^3}

=\sqrt{24}\sqrt{b^3}

=\sqrt{24}b^{\frac{3}{2}}

=\sqrt{24}b^{\frac{3}{2}}\sqrt{40}b^2

=\sqrt{2^3\cdot \:3}b^{\frac{3}{2}}\sqrt{40}b^2

=\sqrt{2^3}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

\sqrt{2^3}

=2^{3\cdot \frac{1}{2}

=2^{3\cdot \frac{1}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3\cdot \:5}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3}\sqrt{5}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}}\sqrt{5}b^2

=\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^2

=\sqrt{3}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^{\frac{3}{2}+2}

2^{\frac{3}{2}+\frac{3}{2}}

=2^3

=2^3\sqrt{3}\sqrt{5}b^{\frac{3}{2}+2}

b^{\frac{3}{2}+2}

=b^{\frac{7}{2}}

=2^3\sqrt{3}\sqrt{5}b^{\frac{7}{2}}

=2^3\sqrt{3\cdot \:5}b^{\frac{7}{2}}

=8\sqrt{15}b^{\frac{7}{2}}

4 0
3 years ago
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