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Minchanka [31]
3 years ago
10

Please help me! Thank you! I only need help with a)

Mathematics
1 answer:
Rzqust [24]3 years ago
8 0

Since we know AD, CD and angle D, we can use the law of cosines, which states that

AC^2 = AD^2+CD^2 - 2AD\cdot CD\cos(D)

Plugging the values, we have

AC^2 = 17^2+13^2 - 2\cdot 17\cdot 13\cos(42)

Which leads to

AC^2 = 458 - 442\cos(42) \approx 129.530

Taking the square root, we have

AC \approx \sqrt{129.530} \approx 11.3811

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I will mark brainliest IF CORRECT
umka2103 [35]

Answer:

1 - (π/6) - (√3/4)

Step-by-step explanation:

First, calculate the area of ABCD. That's just 1*1 = 1 un^{2}

Then, we are told that arcs BD and AC are circular, meaning they are portions of a circle. Therefore, let's redraw this image a little. Look below for the image.

Essentially, what I've done is created an equilateral triangle and two sectors. You may be wondering why the triangle is equilateral. Well, we firstly know it's isosceles because the two original arcs are circular and congruent. Their overlap creates two congruent lines that can be extend from point M to the vertices A and D. These two lines also happen to be radii of the quarter circles because the two sectors have AM AND AD as radii. AM and AD = 1, so AMD is an equilateral triangle.

Draw altitude OD. This is a right angle. Since equilateral triangles are also isosceles triangles, and isosceles triangles have the property of sharing the median to base and altitude to base, we can say that OD = \frac{1}{2} AD = \frac{1}{2}. MD = 1, so this triangle is a 30-60-90 triangle. Why? This is because of the 30º angle converse theorem. The shorter leg is half of the hypotenuse MD, so <OMD is a 30º angle, making <ODM a 60º angle and inevitably <MDC a 30º angle (because <D is a right angle) and <ABM as well because the two sectors are congruent.

Because m<ABM = m<MDC = 30º, these two sectors are 30/360 of the individual circles or 1/12 of the original circles, which have areas of π (because 1^2*π = π). Each is 1/12π, combined they are 1/6π.

The area of the equilateral triangle is √3/8. This is because MO can be calculated, using the 30-60-90 ratios, to be √3/2. (1 * √3/2)/2 = (√3/2)/2 = √3/4; this uses the area of a triangle. Now, time to calculate the shaded area.

The area of the square is 1, the area of the sectors π/6, and the equilateral triangle √3/4. Subtract the equilateral triangle and sectors' areas from the area of the square to be left with the remaining shaded part. 1-(π/6)-(√3/4) is your final answer. To put that a little more cleanly below in the image.

Hope this helps, have a great day.

3 0
2 years ago
Convert 26\%26%26, percent to a fraction in simplest form.
Korolek [52]
26 percent?

A percent is a fraction out of a hundred. 5% is 5 out of one hundred, 20% is 20 out of one hundred, etc.

So 26/100 simplified is 13/50.
6 0
3 years ago
You are building a toothpick fort for class. Each wall must be 1 1⁄3 feet long. Each toothpick is 2 inches long. How many toothp
Mariana [72]

Answer: 68/67.8 toothpicks

Step-by-step explanation

first convert how many inches 11.3 feet is

11.3 = 135.6 inches

so after you get your number you get your calculator and divide it by how many inches the toothpicks are

135.6 ÷ 2= 67.8

then to get a solid answer you would round it to a whole number

67.8--> 68

so you ether say the rounded number or you cut .2 inches off the last toothpick

(also im not sure if its 11.3ft or 113ft so if so then take away the decimal and do the same math, you'll end up with 678 toothpicks)

6 0
3 years ago
A box has dimensions of 17 inches long, 1.3 feet wide, and 8 inches high. What is the volume of the box? The formula for the vol
s344n2d4d5 [400]
I hope this helps you



Volume =17×1,3×8


Volume =176,8
6 0
3 years ago
Simplify the cube root in simplest radical form.<br> 9x374<br> Please help
algol [13]

Answer:

D. xy\sqrt[3]{9y}

Step-by-step explanation:

\sqrt[3]{9x^3y^4}

\sqrt[3]{9}\sqrt[3]{x^3}\sqrt[3]{y^4}

The \sqrt[3]{x^3} cancels out to become x:

\sqrt[3]{9}x\sqrt[3]{y^4}

Split the y^4=y*y^3

\sqrt[3]{9}x\sqrt[3]{y^3}\sqrt[3]{y^1}

\sqrt[3]{y^3} =y

xy\sqrt[3]{9} \sqrt[3]{y}

Put the cube root of y and cube root of 9 together:

xy\sqrt[3]{9y}

6 0
3 years ago
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