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3 years ago

5. Use the rules for long division to divide 262 by 9. A. 29r1 B. 28 C. 30 D. 29

1 answer:

6 0

Since $270=262+8$ , and $\dfrac{270}9=30$ , we have

$\dfrac{270}9=\dfrac{262}9+\dfrac89$
$\implies30-\dfrac89=\dfrac{262}9$
$\implies29+\dfrac19=\dfrac{262}9$

so the answer is A.

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What times what equals 20 but adds up to 19 also ?

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3 years ago

Using the law of cosines, write an algebraic proof to show that the angles in an equilateral triangle must equal 60°. Use "∧" to

bekas [8.4K]

The algebraic proof shows that the angles in an equilateral triangle must equal 60° each

<h3>Laws of cosines </h3>

From the question, we are to use the law of cosines to write an algebraic proof that shows that the angles in an equilateral triangle must equal 60°.

Given any triangle ABC, the measures of angles A, B, and C by the law of cosines are

cos A = (b^2 + c^2 - a^2)/2bc

cos B= (a^2 + c^2 - b^2)/2ac

cos C = (a^2 + b^2 - c^2)/2ab

Now, given that the triangle is equilateral, with each of the side lengths equal to s

That is, a = b = c = s

Then, we can write that

cos A = (s^2 + s^2 - s^2)/(2s×s)

cos A = (s^2 )/(2s^2)

cos A = 1/2

cos A = 0.5

∴ A = cos⁻¹(0.5)

A = 60°

Also

cos B = (s^2 + s^2 - s^2)/(2s×s)

cos B = (s^2 )/(2s^2)

cos B = 1/2

cos B = 0.5

∴ B = cos⁻¹(0.5)

B = 60°

and

cos C = (s^2 + s^2 - s^2)/(2s×s)

cos C = (s^2 )/(2s^2)

cos C = 1/2

cos C = 0.5

∴ C = cos⁻¹(0.5)

C = 60°

Thus,

A = 60°, B = 60° and C = 60°

Hence, the algebraic proof above shows that the angles in an equilateral triangle must equal 60° each.

Learn more on The law of cosines here: brainly.com/question/2866347

#SPJ1

3 0

2 years ago

10 times the sum of a number and 9

Alexandra [31]

Answer:

10(x+9)

Step-by-step explanation:

7 0

4 years ago

Read 2 more answers

Help anyone can help me do 16 and 17 question,I will mark brainlest.The no 16 question is find the area of the shaded region

yanalaym [24]

Answer:

Question 16 = 22

Question 17 = 20 cm²

Step-by-step explanation:

Concepts:

Area of Square = s²

s = side

Area of Triangle = bh/2

b = base
h = height

Diagonals of the square are congruent and bisect each other, which forms a right angle with 90°

Segment addition postulate states that given 2 points A and C, a third point B lies on the line segment AC if and only if the distances between the points satisfy the equation AB + BC = AC.

Solve:

Question # 16

Step One: Find the total area of two squares

Large square: 5 × 5 = 25

Small square: 2 × 2 = 4

25 + 4 = 29

Step Two: Find the area of the blank triangle

b = 5 + 2 = 7

h = 2

A = bh / 2

A = (7) (2) / 2

A = 14 / 2

A = 7

Step Three: Subtract the area of the blank triangle from the total area

Total area = 29

Area of Square = 7

29 - 7 = 22

-----------------------------------------------------------

Question # 17

Step One: Find the length of PT

Given:

PR = 4 cm
RT = 6 cm

PT = PR + RT [Segment addition postulate]

PT = (4) + (6)

PT = 10 cm

Step Two: Find the length of S to PT perpendicularly

According to the diagonal are perpendicular to each other and congruent. Therefore, the length of S to PT perpendicularly is half of the diagonal

Length of Diagonal = 4 cm

4 ÷ 2 = 2 cm

Step Three: Find the area of ΔPST

b = PT = 10 cm

h = S to PT = 2 cm

A = bh / 2

A = (10)(2) / 2

A = 20 / 2

A = 10 cm²

Step Four: Find the length of Q to PT perpendicularly

Similar to step two, Q is the endpoint of one diagonal, and by definition, diagonals are perpendicular and congruent with each other. Therefore, the length of Q to PT perpendicularly is half of the diagonal.

Length of Diagonal = 4 cm

4 ÷ 2 = 2 cm

Step Five: Find the area of ΔPQT

b = PT = 10 cm

h = Q to PT = 2 cm

A = bh / 2

A = (10)(2) / 2

A = 20 / 2

A = 10 cm²

Step Six: Combine area of two triangles to find the total area

Area of ΔPST = 10 cm²

Area of ΔPQT = 10 cm²

10 + 10 = 20 cm²

Hope this helps!! :)

Please let me know if you have any questions

6 0

3 years ago