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user100 [1]
3 years ago
8

Nathan lifts a 5 lb. box 2 feet. How much work has he done?

Mathematics
1 answer:
vekshin13 years ago
7 0

Answer:

10 ft-lbs.

Step-by-step explanation:

Given data

Weight of box=  5 lb

Height moved = 2 feet

We know that the expression for work done is

Work done= Force * Distance

Work done= 5*2

Work done= 10 10 ft-lbs.

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If 6 is a factor of a number, what other numbers must be factors of<br> the number?
zysi [14]

Answer:

2 and 3

Step-by-step explanation:

6 = 2 × 3, so if a number is divisible by 6, it is also divisible by 2 and 3.

8 0
3 years ago
What is the general term equation, an, for the arithmetic sequence 5, 1, â3, â7, . . . , and what is the 21st term of this seque
Valentin [98]
Hello :

the nth term ( general term  of an arithmetic sequence is :

An =A1+(n-1)d   ....a common difference is : d , A1 : the first terme
A1 = 5    and : d = 1 - 5 = - 4    n = 21
A21 = 5 + (21 -1)(-4)
A21 = 5 - 80 = -75

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3 0
3 years ago
Each swing that Susan builds requires $10 in hardware and $20 in wood. She uses the expression 10 + 20 to represent the total co
jenyasd209 [6]
The answer is C. If the original equation is 10+20 for one swing, you multiply that by 3 to get the cost of three swings
7 0
3 years ago
A production process produces 2% defective parts. a sample of 5 parts from the production is selected. what is the probability t
Lana71 [14]

Answer:

Probability of a sample that contains exactly two defective parts is .0037 or .37%

Step-by-step explanation:

As we know if P is the probability of achieving k results in n trials then probability formula is P = \binom{n}{k}p^{K}q^{n-k}

In this formula n = number of trials

                        k = number of success

                        (n-k) = number of failures

                         p = probability of success in one trial

                         q = (1-p) = probability of failure in one trial

In this sum n = 5

                  k = 2

number failures (n-k) = (5-2) = 3

                            p = 2% which can be written as .02

                            q = 98% Which can be written as .98

Now putting these values in the formula

                        P = \binom{5}{2}(.02)^{2}(.98)^{5-2}

                        P = \binom{5}{2}(.02)^{2}(.98)^{3}

                   \binom{5}{2}= 5!/3!2!    

                                              = 5×4×3×2×1/3×2×1×2×1

                                              = 5×2 =10

                                       P = 10×(.02)²×(.98)³

                                          = .0037 or .37%

4 0
3 years ago
Read 2 more answers
1 Factor completely.
bulgar [2K]

1 Factor completely.

p2+7p+10 = (p + 5)(p + 2)



 2 Factor completely.

x2+10x+16 = (x + 4)(x + 6)


3 Factor completely.

x2+7x−18 = (x + 9)(x - 2)



 4 Factor completely.

x2−5x−24 = (x - 8)(x + 3)



 5 Factor completely.

x2−3x−40 = (x - 8)(x + 5)

6 0
3 years ago
Read 2 more answers
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