Asher pays $156 every 6 weeks for piano lessons. What is the price per year for piano lessons

BE

cestrela7 [59]

I don’t see the table necessary in order to solve this question

5 0

2 years ago

Find the indicated real nth roots of a for n=3 ,a=8

Svetach [21]

Answer:

8^ (1/3) =2

Step-by-step explanation:

We want to find cube root of 8

8^ (1/3)

What number times itself 3 times is 8

2*2*2 =8

8^ (1/3) =2

6 0

3 years ago

Please help thank you!

Akimi4 [234]

C. is the answer

.....

3 0

2 years ago

Perform the indicated operations. Reduce the answers to lowest terms .<br>

kozerog [31]

Step-by-step explanation:

1. (2+4+1)/9 = 7/9

2. 2 1/3 + 2/3 = 2 + (1+2)/3 = 2 + 3/3 = 2+1 = 3

3. 1 1/5 + 2 3/5 = 1+2 + 1/5 + 3/5 = 3 + (1+3)/5 =

3 4/5

4. 5/6 + 2/10 + 1/5 = 5/6 + 1/5 + 1/5 = 5/6 + 2/5

= (5×5)/(6×5) + (2×6)/(5×6)

= 25/30 + 12/30

= (25+12)/30

= 37/30 = 1 7/30

5. 3 1/2 + 4 2/3

= 3+4 + 1/2 + 2/3

= 7 + (1×3)/(2×3) + (2×2)/(3×2)

= 7 + 3/6 + 4/6

= 7 + (3+4)/6 = 7 7/6 = 8 1/6

6. 9/13 - 5/13 = (9-5)/13 = 4/13

7. 7 6/8 - 5 2/8

= (7-5) + (6/8 - 2/8)

= 2 + 4/8

= 2 1/2

8. 2/3 - 3/7

= (2×7)/(3×7) - (3×3)/(7×3)

= 14/21 - 9/21 = (14-9)/21 = 5/21

9. 11 1/5 - 5 4/5

= 10 6/5 - 5 4/5

= (10-5) + (6/5 - 4/5)

= 5 + 2/5 = 5 2/5

10. 15 4/5 - 7 7/10

= (15-7) + (4/5 - 7/10)

= 8 + (4×2)/(5×2) - 7/10

= 8 + 8/10 - 7/10

= 8 + 1/10

= 8 1/10

6 0

2 years ago

Prove that a triangle with the sides a - 1 cm to root under a

vivado [14]

Question is Incomplete, Complete question is given below.

Prove that a triangle with the sides (a − 1) cm, 2√a cm and (a + 1) cm is a right angled triangle.

Answer:

∆ABC is right angled triangle with right angle at B.

Step-by-step explanation:

Given : Triangle having sides (a - 1) cm, 2√a and (a + 1) cm.

We need to prove that triangle is the right angled triangle.

Let the triangle be denoted by Δ ABC with side as;

AB = (a - 1) cm

BC = (2√ a) cm

CA = (a + 1) cm

Hence,

${AB}^2 = (a -1)^2$

Now We know that

$(a- b)^2 = a^2+b^2 - 2ab$

So;

${AB}^2= a^2 + 1^2 -2\times a \times1$

${AB}^2 = a^2 + 1 -2a$

Now;

${BC}^2 = (2\sqrt{a})^2= 4a$

Also;

${CA}^2 = (a + 1)^2$

Now We know that

$(a+ b)^2 = a^2+b^2 + 2ab$

${CA}^2= a^2 + 1^2 +2\times a \times1$

${CA}^2 = a^2 + 1 +2a$

${CA}^2 = AB^2 + BC^2$

[By Pythagoras theorem]

$a^2 + 1 +2a = a^2 + 1 - 2a + 4a\\\\a^2 + 1 +2a= a^2 + 1 +2a$

Hence, ${CA}^2 = AB^2 + BC^2$

Now In right angled triangle the sum of square of two sides of triangle is equal to square of the third side.

This proves that ∆ABC is right angled triangle with right angle at B.

4 0

3 years ago