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lions [1.4K]
3 years ago
15

EXTRA POINTS will mark BRAINLIEST! I only need 10, 11 , 16, 18 &19

Mathematics
1 answer:
LuckyWell [14K]3 years ago
6 0

Answer:

10, 11 , 16, 18 &19

Step-by-step explanation:

  • \geq \neq \int\limits^a_b {x} \, dx \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]
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The area of a circle is 78.5 cm² . Find the lenght of the radius of a circle . Use π = 3.14
Tpy6a [65]

Answer:

5cm  = r

Step-by-step explanation:

The area of a circle is given by

A = pi r^2

78.5 = 3.14 r^2

Divide each side by 3.14

78.5/ 3.14 = r^2

25 = r^2

Take the square root of each side

sqrt(25) = sqrt(r^2)

5 = r

6 0
2 years ago
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2. Each of the bracelets that Veronica made has 45 beeds. Shemede 4 bracelets.
cluponka [151]
She used 180 beads because 45x4 =180
6 0
2 years ago
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What number is 10times bigger than 30​
barxatty [35]

Answer:

300

Step-by-step explanation:

30 * 10 = 300

5 0
3 years ago
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Solving Graphically...<br>PLEASE HELP
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Infinitely many? Arrows indicate that the graph goes on continuously without a stopping point, even in parabolas.
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2 years ago
Find two unit vectos that are orthogonal to both [0,1,2] and [1,-2,3]
alekssr [168]

Answer:

Let the vectors be

a = [0, 1, 2] and

b = [1, -2, 3]

( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.

Let the cross product be another vector c.

To find the cross product (c) of a and b, we have

\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]

c = i(3 + 4) - j(0 - 2) + k(0 - 1)

c = 7i + 2j - k

c = [7, 2, -1]

( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:

c / | c |

Where | c | = √ (7)² + (2)² + (-1)²  = 3√6

Therefore, the unit vector is

\frac{[7,2,-1]}{3\sqrt{6} }

or

[ \frac{7}{3\sqrt{6} } , \frac{2}{3\sqrt{6} } , \frac{-1}{3\sqrt{6} } ]

The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:

[ \frac{-7}{3\sqrt{6} } , \frac{-2}{3\sqrt{6} } , \frac{1}{3\sqrt{6} } ]

In conclusion, the two unit vectors are;

[ \frac{7}{3\sqrt{6} } , \frac{2}{3\sqrt{6} } , \frac{-1}{3\sqrt{6} } ]

and

[ \frac{-7}{3\sqrt{6} } , \frac{-2}{3\sqrt{6} } , \frac{1}{3\sqrt{6} } ]

<em>Hope this helps!</em>

7 0
2 years ago
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