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3 years ago

Can you help me please

Mathematics

1 answer:

Readme [11.4K]3 years ago

5 0

The area of the sector which is the white triangle adding the shaded region is 68.9/360*π*9.28^2=51.78005(rounding to 5th digit after decimal point for accuracy before we do final round for answer)

The area of the white triangle in the sector has area 1/2*9.28^2*sin(68.9)= 40.17223(rounding to 5 digits again for some accuracy.

Now we take out the white triangle from the sector.

51.78005-40.17223=11.60782

rounding to the nearest tenth we get 11.6 cm^2

Problem done!

Hope this helped and if you have any questions about my explanation just ask in the comment and I will answer.

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SAVE ME. Please Answer my Questions Clevers.

Vladimir [108]

Answer:

See below

Step-by-step explanation:

Q1. if a counting number is selected at random from a set of whole number less than or equal to 20 find the probability of getting:

Counting numbers are Natural numbers: $\mathbb{N}$

Also, we have Whole numbers. Despite not having an official symbol, I usually denote the set as $\mathbb{Z}_{\ge 0}$

Whole numbers less than or equal to 20: $A\leq 20, A \subset \mathbb{Z}_{\ge 0} \\\implies A=\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}$

A. Prime numbers

From the set A, the prime numbers are 2, 3, 5, 7, 11, 13, 17, 19.

Once we have 21 numbers in total and 8 prime numbers, the probability is:

$P=\frac{8}{21} \approx 40\%$

B. Composite numbers

From the set A, the composite numbers are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20.

Once we have 21 numbers in total and 11 composite numbers, the probability is:

$P=\frac{11}{21} \approx 52\%$

C. Divisible by three

From the set A, the numbers divisible by three are 3, 6, 9, 12, 15, 18.

Once we have 21 numbers in total and 6 numbers divisible by three, the probability is:

$P=\frac{6}{21} \approx 30\%$

D. Square of 2

From the set A, the numbers square of 2 are 0, 1, 4, 9, 16.

$\sqrt{0} =0$

$\sqrt{1} =1$

$\sqrt{4} =2$

$\sqrt{9} =3$

$\sqrt{16} =4$

Once we have 21 numbers in total and 5 numbers square of 2 , the probability is:

$P=\frac{5}{21} \approx 24\%$

Q2. The opposite angle of acyclic quadrilaterals is in the ratio of 2:3. Find the degree measure of each opposite angle?

Once they're opposite, they add up to 180º

$2x+3x=180 \implies 5x=180 \implies x=36$

The first angle is 72º

The second angle is 108º

Q3. what is the solution set of 9x-4<13x-7 (domain, x e z) ?

$x \in \mathbb{Z}\\$

$x>\frac{3}{4}$

$x\in \left(\frac{3}{4},\infty \right)$

Q4. if three fourth of a number is one tenths, what is the number?

$\frac{3}{4} x =\frac{1}{10} \implies 3x=\frac{4}{10} \implies \boxed{x = \frac{2}{15}}$

Q5. which one is the equation of the line passing through the origin and having a slope 4?

$y=mx+b$

$m: \text{slope}$

$b: \text{y-intercept}$

B. Y= 4x

7 0

3 years ago

What are the coordinates of the point on the directed line segment from (–9, -7) to

pentagon [3]

Answer:

Step-by-step explanation:

Saving the long, drawn out derivation of the formulas to find the x and y coordinates of the directed point, suffice it to say that it is:

x coordinate: $\frac{bx_1+ax_2}{a+b}$ and

y coordinate: $\frac{by_1+ay_2}{a+b}$

where x1, x2, y1, and y2 are the coordinates from the given points and a and b are the numbers in the ratio, namely a = 3 and b = 4. Filling in accordingly:

the x coordinate of the directed point is

$\frac{4(-9)+3(-2)}{7}$ which simplifies down to -6, and

the y coordinate of the directed point is

$\frac{4(-7)+3(7)}{7}$ which simplifies down to -1.

The coordinate of the point is (-6, -1). Write that down so you don't forget it.

4 0

3 years ago

Select all the conditions for which it is possible to construct a triangle. (7.G.1.2) Group of answer choices a. A triangle with

saw5 [17]

Answer:

b, d, e, f

Step-by-step explanation:

Here are the applicable restrictions:

The sum of angles in a triangle is 180°, no more, no less.
The sum of the lengths of the two shortest sides exceeds the longest side.
When two sides and the angle opposite the shortest is given, the sine of the given angle must be at most the ratio of the shortest to longest sides.

a. A triangle with angle measures 60°, 80°, and 80° (angle sum ≠ 180°, not OK)

b. A triangle with side lengths 4 cm, 5 cm, and 6 cm (4+5 > 6, OK)

c. A triangle with side lengths 4 cm, 5 cm, and 15 cm (4+5 < 15, not OK)

d. A triangle with side lengths 4 cm, 5 cm, and a 50° angle across from the 4 cm side (sin(50°) ≈ 0.766 < 4/5, OK)

e. A triangle with angle measures 30° and 60°, and an included 3 cm side length (OK)

f. A triangle with angle measures 60°, 20°, and 100° (angle sum = 180°, OK)

_____

Additional comment

In choice "e", two angles and the side between them are specified. As long as the sum of the two angles is less than 180°, a triangle can be formed. The length of the side is immaterial with respect to whether a triangle can be made.

The congruence postulates for triangles are ...

SSS, SAS, ASA, AAS, and HL

These essentially tell you the side and angle specifications necessary to define a singular triangle. As we discussed above, the triangle inequality puts limits on the side lengths specified in SSS. The angle sum theorem puts limits on the angles when only two are specified (ASA, AAS).

In terms of sides and angles, the HL postulate is equivalent to an SSA theorem, where the angle is 90°. In that case, the angle is opposite the longest side (H). In general, SSA will specify a singular triangle when the angle is opposite the longest specified side, regardless of that angle's measure. However, when the angle is opposite the shortest specified side, the above-described ratio restriction holds. If the sine of the angle is less than the ratio of sides, then two possible triangles are specified.

4 0

2 years ago

Please help me , i just need to past this !

OLga [1]

Answer:

Step-by-step explanation:

i would say c

3 0

3 years ago

Read 2 more answers

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Maurinko [17]

I think it is $30 since it has the highest probability.

7 0

3 years ago

Can you help me please ​

Can you help me please