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lbvjy [14]
3 years ago
5

two forces of magnitude 100n and 70n act on an object. the angle between the forces is 120 degrees. what is the resultant force?

Mathematics
1 answer:
Katen [24]3 years ago
5 0
The resultant force would be 88.88 N at an angle of 43 degrees from the 100 N force.
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154×35= can you show work
DochEvi [55]
The way to solve it is to put the 154 on top of 35 and multiply it to get the answer, 154 times 35 = 5390
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3 years ago
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9 - 10x = 2x + 1 - 8x
Furkat [3]

Answer:

x=2

Step-by-step explanation:

9 - 10x = 2x + 1 - 8x

9-10x=-6x+1

Subtract 9 from both sides:

9-10x-9=-6x+1-9

-10x=-6x-8

Add 6x to both sides:

-10x+6x=-6x-8+6x

-4x=-8

Divide both sides by -4:

\frac{-4x}{-4}=\frac{-8}{-4}

x=2

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3 years ago
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Use a factor and zero product property to solve the following equation 6x(x+4)=0
Rina8888 [55]
6x(x+4)=0 \\
6x=0 \ \lor \ x+4=0 \\
x=0 \ \lor \ x=-4 \\
\boxed{x=0 \hbox{ or } x=-4}
4 0
2 years ago
Write the addition statement represented by each diagram <br><br> -8 <br> +5
solmaris [256]

Answer:

Here is the diagram that Han drew to represent 0.25. Draw a different diagram that represents 0.25. Explain why your diagram and Han's diagram represent the same number. Figure \(\Page Index{9}\) For each of these numbers, draw or describe two different diagrams that represent it. \(0.1\) \(0.02\) \(0.43\) Use diagrams of base-ten units to.

Step-by-step explanation:

3 0
2 years ago
The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typo
muminat

Answer:

The required probability is 0.55404.

Step-by-step explanation:

Consider the provided information.

The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typographical errors on a page of second booklet is a Poisson random variable with mean 0.3.

Average error for 7 pages booklet and 5 pages booklet series is:

λ = 0.2×7 + 0.3×5 = 2.9

According to Poisson distribution: {\displaystyle P(k{\text{ events in interval}})={\frac {\lambda ^{k}e^{-\lambda }}{k!}}}

Where \lambda is average number of events.

The probability of more than 2 typographical errors in the two booklets in total is:

P(k > 2)= 1 - {P(k = 0) + P(k = 1) + P(k = 2)}

Substitute the respective values in the above formula.

P(k > 2)= 1 - ({\frac {2.9 ^{0}e^{-2.9}}{0!}} + \frac {2.9 ^{1}e^{-2.9}}{1!}} + \frac {2.9 ^{2}e^{-2.9}}{2!}})

P(k > 2)= 1 - (0.44596)

P(k > 2)=0.55404

Hence, the required probability is 0.55404.

4 0
2 years ago
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