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2 years ago

Find the length of each arc. Round to the nearest tenth.

Mathematics

1 answer:

avanturin [10]2 years ago

7 0

Answer:

arc ≈ 5.5 units

Step-by-step explanation:

The arc of the circle is calculated as

arc = circumference of circle × fraction of circle

= 2πr × $\frac{\frac{\pi }{2} }{2\pi }$ ( r is the radius )

= 2π × 3.5 × $\frac{1}{4}$

= $\frac{7\pi }{4}$

≈ 5.5 ( to the nearest tenth )

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calculeaza lungimea segmentului ab in fiecare dintre cazuri:A(1,5);B(4,5);A(2,-5),B(2,7);A(3,1)B(-1,4);A(-2,-5)B(3,7);A(5,4);B(-

Tatiana [17]

Answer:

1. 3; 2. 12; 3. 5; 4. 13; 5. 10; 6. 10

Step-by-step explanation:

We can use the distance formula to calculate the lengths of the line segments.

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

1. A (1,5), B (4,5) (red)

$d = \sqrt{(x_{2} - x_{1}^{2}) + (y_{2} - y_{1})^{2}} = \sqrt{(4 - 1)^{2} + (5 - 5)^{2}}\\= \sqrt{3^{2} + 0^{2}} = \sqrt{9 + 0} = \sqrt{9} = \mathbf{3}$

2. A (2,-5), B (2,7) (blue)

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(2 - 2)^{2} + (7 - (-5))^{2}}\\= \sqrt{0^{2} + 12^{2}} = \sqrt{0 + 144} = \sqrt{144} = \mathbf{12}$

3. A (3,1), B (-1,4 ) (green)

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(-1 - 3)^{2} + (4 - 1)^{2}}\\= \sqrt{(-4)^{2} + 3^{2}} = \sqrt{16 + 9} = \sqrt{25} = \mathbf{5}$

4. A (-2,-5), B (3,7) (orange)

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(3 - (-2))^{2} + (7 - (-5))^{2}}\\= \sqrt{5^{2} + 12^{2}} = \sqrt{25 + 144} = \sqrt{169} = \mathbf{13}$

5. A (5,4), B (-3,-2) (purple)

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(-3 - 5)^{2} + (-2 - 4)^{2}}\\= \sqrt{(-8)^{2} + (-6)^{2}} = \sqrt{64 + 36} = \sqrt{100} = \mathbf{10}$

6. A (1,-8), B (-5,0) (black)

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(-5 - 1)^{2} + (0 - (-8))^{2}}\\-= \sqrt{(-6)^{2} + (-8)^{2}} = \sqrt{36 + 64} = \sqrt{100} = \mathbf{10}$

6 0

3 years ago

Solve 10 over 8=n over 10

Vikki [24]

10/8 = n/10

First, we need to simplify 10/8 into the lowest terms it can go down into. To do so, we need to find the greatest common factor (GCF) of both the numerator and denominator (10 and 8).

Factors of 10: 1, 2, 5, 10
Factors of 8: 1, 2, 4, 8

Looking at the listed factors above for our two numbers, we can see that the greatest common factor is 2. The GCF is 2.

Second, our next step for simplifying 10/8 down is to divide the numerator (10) and denominator (8) by our GCF we recently found which was 2.

$10 \div 2 = 5 \\ 8 \div 2 = 4$

We can now rewrite our fraction in the simplified form which is 5/4.

Third, our goal from the start is to get the variable (n) to one side of the problem by itself. This means we have to do everything else on the other side of the problem. We can start out by multiplying each side by 10.
$\frac{5}{4} \times 10 = n$

Fourth, we now need to simplify 5/4 times 10. To do this, we take the numerator and multiply it by 10. Since the numerator is 5, our problem should look like: 5 × 10. The answer is 50.
$\frac{50}{4} = n$

Fifth, since 50/4 can be simplified as well into lower terms. Let's do the same thing we did earlier when we simplified the earlier fraction. List the factors of the numerator (50) and denominator (4).

Factors of 50: 1, 2, 5, 10, 25, 50
Factors of 4: 1, 2, 4

Out of the factors for both of the numbers, which are common? 1 and 2 are the common factors and 2 is the greatest, meaning that 2 is our GCF.

Sixth, like we did earlier as well, we now have to divide our numerator and denominator by our recently found GCF which is 2.

$50 \div 2 = 25 \\ 4 \div 2 = 2$

We now have our simplified fraction, which is also our answer. The simplified fraction is 25/2.

Answer in fraction form: $\fbox {n = 25/2}$
Answer in decimal form: $\fbox {x = 12.5}$

6 0

3 years ago

Amy is using a drawing program to complete a construction with which she is almost finished. Which construction is she completin

Vedmedyk [2.9K]

Given options : Two intersecting circles are drawn with a radius in each marked. the image will be linked.

Given options : An equilateral triangle inscribed in a circle

A square inscribed in a circle

A regular pentagon inscribed in a circle

A regular hexagon inscribed in a circle.

<u>Note. When we join an intersection point of two circles and centers of the circles it would form an equilateral triangle that would be inscribe inside a common portion of both circles..</u>

Therefore, an equilateral triangle inscribed in a circle would be correct option.

She is completing an equilateral triangle inscribed in a circle.