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4 years ago

A violinist has practiced 15 pieces. how many different combinations are possible if he chooses 6 of these pieces for a recital?

1 answer:

8 0

There are 3603600 possible combinations

this can be found by multiplying:
1/15,1/14,1/13,1/12,1/11,1/10 together and using the denominator as the final value

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Pleaseee helppppppppppppppppppp

quester [9]

To find Angle A we use cosine

cos ∅ = adjacent / hypotenuse

From the question

The adjacent is 17

The hypotenuse is 38

So we have

cos A = 17/38

A = cos-¹ 17/38

A = 63.4

<h3>A = 63° to the nearest degree</h3>

To find Angle C we use sine

sin ∅ = opposite / hypotenuse

From the question

The opposite is 17

The hypotenuse is 38

So we have

sin C = 17/38

C = sin-¹ 17/38

C = 26.57

<h3>C = 27° to the nearest degree</h3>

Hope this helps you

7 0

3 years ago

ishmal invested £3500 for 3 years at 2.5% per annum simple interest. Work out the total amount of interest Ishmal earned.

Jlenok [28]

Answer:

Interest Earned is $262.5

Step-by-step explanation:

Interest=PRT/100

When P=Principal, R=Rate, T=Time

Interest earn =( 3500 x 3 x 2.5)/ 100

I=$26,250 / 100

I=$262.5

6 0

4 years ago

The simplified form of an expression is 1/258 t^28. Which expression was simplified

Aleksandr-060686 [28]

<span>1/258 *(t^28)
= t^28 / 4^4
= t^28 4^-4
= (t^-7 * 4)^-4
= (4t^-7)^-4

answer is D. last option</span>

7 0

3 years ago

Read 2 more answers

Independent Practice

motikmotik

The answer you’re looking for is a yes the domain value five corresponds to two range values -8 and five

5 0

3 years ago

Please find the perimeter and area of these shapes

ladessa [460]

Answer:

ABC shaded area = 36 $\pi$ - 72 cm²

ABC shaded area perimeter = $6\pi +12\sqrt{2}$ cm

ABCD area = $\dfrac52 \pi$ cm²

ABCD perimeter = $3\pi +2$ cm

Step-by-step explanation:

<u>Shape ABC</u>

Assuming you want the area and perimeter of the shaded part of the shape only...

Area of a sector = $\dfrac12r^2\theta$ (where r is the radius and $\theta$ <em> </em>

⇒ area of a sector = $\dfrac12 \times 12^2\times \dfrac{\pi}{2} =36\pi \ \textsf{cm}^2$

Area of triangle = 1/2 x base x height

⇒ area of triangle = 1/2 x 12 x 12 = 72 cm²

Therefore, area of shaded area = area of sector - area of triangle

⇒ area = 36 $\pi$ - 72 cm²

<u>Perimeter</u>

Arc length = $r\theta$ (where r is the radius and $\theta$ <em> </em>

⇒ arc length = $12\times\dfrac12\pi =6\pi \ \textsf{cm}$

Hypotenuse of triangle = $\sqrt{a^2+b^2}$ (where a and b are the legs of the right triangle)

⇒ hypotenuse = $\sqrt{12^2+12^2} =12\sqrt{2}$ cm

Therefore, perimeter = arc length + hypotenuse

⇒ perimeter = $6\pi +12\sqrt{2}$ cm

<u>Shape ABCD</u>

Area of a semicircle = $\dfrac12 \pi r^2$ (where r is the radius)

⇒ area of large semicircle ABC = $\dfrac12 \times \pi \times 2^2=2\pi \ \textsf{cm}^2$

⇒ area of small semicircle AD = $\dfrac12 \times \pi \times 1^2=\dfrac12\pi \ \textsf{cm}^2$

⇒ area of shape ABCD = $\dfrac12 \pi + 2 \pi=\dfrac52 \pi \ \textsf{cm}^2$

<u>Perimeter</u>

1/2 circumference = $\pi r$

⇒ perimeter = $2\pi +2+\pi=3 \pi+2 \ \textsf{cm}$

7 0

2 years ago