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Debora [2.8K]
3 years ago
11

To make each costume Rachel uses 6 yards of material and 3 yards of trim supposed to use as a total of 48 yards of material to m

ake several costumes how many yards of trim does she use right to reuse your friend the number of yards of trim
Mathematics
1 answer:
pochemuha3 years ago
7 0

Answer:

24 yards of trim

Step-by-step explanation:

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Gala2k [10]

Answer:

5x -11

Step-by-step explanation:

it says it there

5 0
3 years ago
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I need help on this Question.!
OLga [1]

Answer:

see below

Step-by-step explanation:

m/n = 1/7

Using cross products

7m = n

This is a direct proportion

7 0
3 years ago
Help me with these please​
olchik [2.2K]

Step-by-step explanation:

(1) y = x e^(x²)

Take derivative with respect to x:

dy/dx = x (e^(x²) 2x) + e^(x²)

dy/dx = 2x² e^(x²) + e^(x²)

dy/dx = (2x² + 1) e^(x²)

Take derivative with respect to x again:

d²y/dx² = (2x² + 1) (e^(x²) 2x) + (4x) e^(x²)

d²y/dx² = (4x³ + 2x) e^(x²) + 4x e^(x²)

d²y/dx² = (4x³ + 6x) e^(x²)

Substitute:

d²y/dx² − 2x dy/dx − 4y

= (4x³ + 6x) e^(x²) − 2x (2x² + 1) e^(x²) − 4x e^(x²)

= 4x³ + 6x − 2x (2x² + 1) − 4x

= 4x³ + 6x − 4x³ − 2x − 4x

= 0

(2) y = sin⁻¹(√x)

sin y = √x

sin²y = x

Take derivative with respect to x:

2 sin y cos y dy/dx = 1

sin(2y) dy/dx = 1

dy/dx = csc(2y)

Take derivative with respect to x again:

d²y/dx² = -csc(2y) cot(2y) 2 dy/dx

d²y/dx² = -2 csc²(2y) cot(2y)

Substitute:

2x (1 − x) d²y/dx² + (1 − 2x) dy/dx

= 2 sin²y (1 − sin²y) (-2 csc²(2y) cot(2y)) + (1 − 2 sin²y) csc(2y)

Use power reduction formula:

= (1 − cos(2y)) (1 − ½ (1 − cos(2y))) (-2 csc²(2y) cot(2y)) + (1 − (1 − cos(2y))) csc(2y)

= (1 − cos(2y)) (1 − ½ + ½ cos(2y)) (-2 csc²(2y) cot(2y)) + cos(2y) csc(2y)

= (1 − cos(2y)) (½ + ½ cos(2y)) (-2 csc²(2y) cot(2y)) + cot(2y)

= (cos(2y) − 1) (1 + cos(2y)) csc²(2y) cot(2y) + cot(2y)

= (cos²(2y) − 1) csc²(2y) cot(2y) + cot(2y)

= -sin²(2y) csc²(2y) cot(2y) + cot(2y)

= -cot(2y) + cot(2y)

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8 0
3 years ago
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igomit [66]

Answer: -18v+9w+15

Step-by-step explanation:

1. -3(6v)-3(-3w)-3(-5)

2.-18v+9w+15

3 0
3 years ago
If the 9th term of an A.P is 0, prove that its 29th term is twice its 19th term
Ghella [55]

Let the first term, common difference and number of terms of an AP are a, d and n respectively.

Given that, 9th term of an AP, T9 = 0 [∵ nth term of an AP, Tn = a + (n-1)d]

⇒ a + (9-1)d = 0

⇒ a + 8d = 0 ⇒ a = -8d ...(i)

Now, its 19th term , T19 = a + (19-1)d

= - 8d + 18d [from Eq.(i)]

= 10d ...(ii)

and its 29th term, T29 = a+(29-1)d

= -8d + 28d [from Eq.(i)]

= 20d = 2 × T19

Hence, its 29th term is twice its 19th term

6 0
3 years ago
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