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frutty [35]
3 years ago
7

A manufacturer makes shipments of building block sets that should ideally have 315 pieces. A quality-control inspector randomly

selects a set. Any set that varies from the ideal number of pieces by more than 8 blocks is sent back. What is the range of allowable number of blocks for a building set?
Mathematics
1 answer:
Elis [28]3 years ago
8 0

Answer:

307 to 323 blocks

Step-by-step explanation:

The range of allowable number of blocks for a building set is the minimum numbers of blocks in a set to the maximum number of blocks in set that would not cause the set to be sent back. Since the ideal number 315 pieces and the allowed deviance is 8 blocks less or more, the allowable range is:

315 - 8 ≤ N ≤ 315 +8

307  ≤ N ≤ 323.

The range of allowable number of blocks for a building set is 307 to 323 blocks.

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y=-3x+3

Step-by-step explanation:

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A cold water tap can fill a tub in 6 minutes, and a hot water tap can fill the tub in 8 minutes. A drain can empty a full tub in
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12 minutes

Step-by-step explanation:

The Hot tap fills 1/30th of the bath in 1 minute. The cold tap fills 1/20th of the bath in one minute. In one minute both taps together will fill (1/30 + 1/20) of the bath, which is 5/60ths or 1/12th of the bath. Since it takes one minute to fill 1/12th of the bath, it’ll take 12 minutes to fill the bath

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3 years ago
Y=2x
Igoryamba

Answer:

( 10,20 )

Step-by-step explanation:

Y=2x

Y= 3x -10

What is the solution to the system of equations?

2x=3x-10      *combine like terms*

-1x= -10 x=10 *divide*

Y=2(10)  y=20  *plug in for y into other equation*

CHECK:

plug both x and y into both equations to get a true statement.

Y=2x   20=2(10)  20=20

Y= 3x -10      20= 3(10) -10  20=20  

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Find the zeros the polynomial fx=2x^2 +9x-5<br>pls help with steps ​
lara [203]

Answer:

-5 ; 1/2

Step-by-step explanation:

<u>GIVEN :-</u>

  • A quadratic polynomial f(x) = 2x² + 9x - 5

<u>TO FIND :-</u>

  • Zeroes of f(x) = 2x² + 9x - 5

<u>GENERAL CONCEPTS TO BE USED IN THIS QUESTION :-</u>

Lets say there's a quadratic polynomial f(x) = ax² + bx + c , whose factors are (x - α) & (x - β). To find the values of x for which f(x) will be zero , equate the factors of f(x) with 0.

⇒ (x - α) = 0 & (x - β) = 0

⇒ x = α & x = β

Hence, it can be concluded that if (x - α) & (x - β) are factors of f(x) , then α & β are the roots of f(x).

<u>SOLUTION :-</u>

Factorise f(x) = 2x² + 9x - 5.

  • Split its middle term.

=> 2x^2 + 10x - x - 5

  • Take '2x' common from first two terms & '-1' from last two terms.

=> 2x(x + 5) - 1(x + 5)

  • Take (x + 5) common from the whole expression.

=> (x + 5)(2x - 1)

So , the factors of f(x) are (x + 5) & (2x - 1). Now equate the factors with zero.

⇒ (x + 5) = 0 & (2x - 1) = 0

⇒ x = -5 & x = 1/2

∴ The zeroes of f(x) = 2x² + 9x - 5 are (-5) & (1/2)

<u>VERIFICATION :-</u>

1) Put x = -5 in f(x) = 2x² + 9x - 5

⇒ f(-5) = 2(-5)² + 9×(-5) - 5

           = 50 - 45 - 5

           = 50 - 50  

           = 0

2) Put x = 1/2 in f(x) = 2x² + 9x - 5

f(\frac{1}{2} ) = 2 \times (\frac{1}{2} )^2 + 9 \times \frac{1}{2}  - 5

       = 2 \times \frac{1}{4} + \frac{9}{2} - 5

       = \frac{1}{2} + \frac{9}{2} - 5

       = \frac{ 1 + 9 - 10}{2}

       = \frac{0}{2}

       = 0

6 0
3 years ago
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