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kirill115 [55]
3 years ago
13

If A=(0,0) and B=(6,3), what is the length of AB ?

Mathematics
2 answers:
motikmotik3 years ago
4 0
D(AB) = <span>√(6-0)^2 +(3-0)^2
d(AB) = </span><span>√36+9
d(AB) = </span><span>√45
d(AB) = 3</span><span>√5
or
d(AB) = 6.71

hope it helps</span>
lozanna [386]3 years ago
4 0

Answer:

6.71 units.

Step-by-step explanation:

We have been given coordinates of two points A(0,0) and B(6,3). We are asked to find the length of AB.

To find the length of Ab, we will use distance formula.

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Let point (0,0)=(x_1,y_1) and (6,3)=(x_2,y_2).

Upon substituting coordinates of our given points in distance formula, we will get:

d=\sqrt{(6-0)^2+(3-0)^2}

d=\sqrt{(6)^2+(3)^2}

d=\sqrt{36+9}

d=\sqrt{45}

d=\sqrt{5\cdot 9}

d=\sqrt{5\cdot 3^2}

d=3\sqrt{5}

d=6.7082039\approx 6.71

Therefore, the length of AB is approximately 6.71 units.

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<u>Answer:</u>

The correct answer option is H. 4374.

<u>Step-by-step explanation:</u>

We are given the following geometric sequence and we are to find its 8th term:

\frac{2}{9}, \frac{2}{3}, 2,...

Here a_1=\frac{2}{9} and common ratio (r) = \frac{\frac{2}{3} }{\frac{2}{9} } =3.

The formula we will use to find the 10th term is:

nth term = a_1 \times r^{(n-1)}

Substituting the values in the formula to get:

10th term = \frac{2}{9} \times 3^{(10-1)}

10th term = 4374

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The product of two positive numbers is 120. One number is 7 more than the other. Determine the two numbers
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If you have 36 ft of fencing what are the area of the different rectangles you could enclose with the fencing?consider only whol
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With 36 feet of fence, these are the rectangles that you can enclose:

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6 0
3 years ago
Read 2 more answers
A bag contains 4 red balls, 2 green balls, 3 yellow balls, and 5 blue balls. Find each probability for randomly removing balls w
AfilCa [17]

Answer:

\frac{15}{4802}, \frac{15}{9604}, \frac{9}{2401}, \frac{9}{4802}

Step-by-step explanation:

The bag has a total of (4+2+3+5) = 14 balls. Set up the proportions:

Red: \frac{4}{14}

green: \frac{2}{14}

yellow: \frac{3}{14}

blue: \frac{5}{14}

Now solve!

Removing 1 yellow, 1 red, 1 green, and 1 blue = \frac{3}{14} \cdot \frac{4}{14}  \cdot \frac{2}{14}  \cdot \frac{5}{14} =\frac{15}{4802}

Removing 1 blue, 1 green, 1 green, and 1 yellow = \frac{5}{14} \cdot \frac{2}{14}  \cdot \frac{2}{14}  \cdot \frac{3}{14} =\frac{15}{9604}

Removing 1 red, 1 red, 1 yellow, and 1 yellow = \frac{4}{14} \cdot \frac{4}{14}  \cdot \frac{3}{14}  \cdot \frac{3}{14} =\frac{9}{2401}

Removing 1 green, 1 yellow, 1 yellow, and 1 red = \frac{2}{14} \cdot \frac{3}{14}  \cdot \frac{3}{14}  \cdot \frac{4}{14} =\frac{9}{4802}

4 0
2 years ago
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