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PolarNik [594]
3 years ago
10

C+4c=15 what’s the answer?

Mathematics
1 answer:
Fynjy0 [20]3 years ago
4 0
C=3 that’s the answer I think
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Gavin was thinking of a number. Gavin halves the number and gets an answer of 34.3. Form an equation with x as information.
Lostsunrise [7]

Gavin's number is 68.6

Equation: 1/2(x) = 34.3

Use inverse operations on both sides.

1/2(x) = 1x or x = 34.3(2)

34.3(2) = 68.6

Thus, x equals 68.6

6 0
3 years ago
What is the answer to 7(2x + 9)
Damm [24]

Answer:

14x + 63

Step-by-step explanation:

7(2x+9)\\\\14x+63\\\\\left \{ {{7*2x=14x} \atop {7*9=63}} \right.

Hope this helps.

7 0
3 years ago
Company 1: 3.4 defects per 980,000 occurrences Company 2: 4 defects per 1,115,783 occurrences Company 3: 18 defects per 5,999,52
Gelneren [198K]

Given Information:  

Company 1: defects = 3.4

Company 1: units = 980,000

Company 2: defects = 4

Company 2: units = 1,115,783

Company 3: defects = 18

Company 3: units = 5,999,521

Required Information:  

six-sigma level = ?

Answer:

Company 1: Possible

Company 3: Impossible

Company 3: Possible

Step-by-step explanation:

The six sigma levels are from 1 to 6 where 1 represents lowest quality level and 6 represents highest quality level

To find out sigma level first we have to find out DPU and DMPU

Where DPU is no. of defects divided by no. of units

And DMPU is DPU multiplied by 1 million

Company 1:

DPU = 3.4/980,000 = 0.0000030612

DMPU = 0.0000030612*1,000,000 = 3.06

Company 2:

DPU = 4/1,115,783 = 0.000003584

DMPU = 0.000003584*1,000,000 = 3.58

Company 3:

DPU = 18/5,999,521 = 0.00000300

DMPU = 0.00000300*1,000,000 = 3.0

According to six sigma tables, for six-sigma level, DMPU must be equal or less than 3.4. As you can see only company 1 and 3 has DMPU less than 3.4 therefore, it is possible for company 1 and company 3 to meet six-sigma level but impossible for company 2.

3 0
3 years ago
PLEASE ANSWERRRRRRRRR
egoroff_w [7]

Answer:

3/2

Step-by-step explanation:

7 0
4 years ago
Read 2 more answers
What are the last three terms of the binomial expansion 32x³ + 80x'y + 80x³y² +
Gnesinka [82]

The last three terms of the given binomial expansion are 40x²y³ + 10xy⁴ + y⁵

<h3>Binomial expansion</h3>

From the question, we are to determine the last three terms of the binomial expansion 32x⁵ + 80x⁴y + 80x³y² + __ + __ + __

The highest power of the expansion is 5

Using the binomial expansion theorem

(m +n)⁵ = m⁵ + 5m⁴n + 10m³n² + 10m²n³ + 5mn⁴ + n⁵

Now, we will determine the values of m and n by comparison

By comparison,

m⁵ = 32x⁵

m⁵ = (2x)⁵

∴ m = 2x

Also,

5m⁴n = 80x⁴y

Divide both sides by 5, to get

m⁴n = 16x⁴y

Put m = 2x into the equation

(2x)⁴n = 16x⁴y

16x⁴n = 16x⁴y

∴ n = y

Thus, the expression being expanded is (2x + y)⁵

By binomial expansion, the expansion of (2x + y)⁵  is

(2x + y)⁵ = (2x)⁵ + 5(2x)⁴y + 10(2x)³(y)² + 10(2x)²(y)³ + 5(2x)(y)⁴ + y⁵

(2x + y)⁵ = 32x⁵ + (5×16)x⁴y + (10×8)x³(y)² + (10×4)x²(y)³ + (5×2)x(y)⁴ + y⁵

(2x + y)⁵ = 32x⁵ + 80x⁴y + 80x³y² + 40x²y³ + 10xy⁴ + y⁵

Hence, the last three terms of the given binomial expansion are 40x²y³ + 10xy⁴ + y⁵

Learn more on Binomial expansion here: brainly.com/question/13602562

#SPJ1

8 0
2 years ago
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