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kondaur [170]
3 years ago
12

Hue is arranging chairs.She can form 6 rows of a given length with 3 chairs left over,or 8 rows of that same length if she gets

11 more chairs.Write and solve an equation to find how many chairs are in that row length
Mathematics
1 answer:
Step2247 [10]3 years ago
5 0

Answer:

There are 7 chairs in each row.

Step-by-step explanation:

Hue is arranging some number of chairs. If she arranges them in 6 rows of equal lengths then there will be 3 chairs leftover, or she arranges them in 8 rows of that same length then she requires 11 more chairs.

Let us assume that there are P numbers of chairs and there are x chairs in each row.

Therefore, we can write that  

6x + 3 = P ....... (1) and  

8x = P + 11  

⇒ 8x - 11 = P ........ (2)

Now, from equations (1) and (2) we get,

6x + 3 = 8x - 11

⇒ 2x = 14

⇒ x = 7

Therefore, there are 7 chairs in each row. (Answer)

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Hello : 
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3 years ago
Find the area of the unshaded region for these two problems​
Grace [21]

1. Area of the right angled ∆ = 1/2 × base × height

height = 16m

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2 years ago
A coin is tossed four times What is the probability of getting four heads? (1 mark) a) b) What is the probability of getting exa
BlackZzzverrR [31]
<h2>Answer:</h2>

1. A coin is tossed four times. What is the probability of getting four heads?

Each toss has a 1/2 chance of getting a head.

So the chance of getting all four heads can be calculated as :

1/2\times1/2\times1/2\times1/2=1/16

2. A coin is tossed four times. What is the probability of getting two heads?

Each toss can have 2 results, so 4 flips will have 2^{4} or 16 results.

Getting two heads means getting two tails also. So, we can get the number of times two heads come = \frac{4!}{2!2!} = 6

We can write the groups like - {HHTT,HTHT,HTTH,THHT,THTH,TTHH}

So, the required probability is : 6/16 or 3/8.

3. Not getting two heads means getting 3 tails and 1 head or all tails.

Probability of having all tails = 1/2\times1/2\times1/2\times1/2=1/16

Probability of one head(1st trial) and three tails = 1/16

Probability of one head (2nd trial) and three tails (the 1st, 3rd and 4th trials) = 1/16

Probability of one head (3rd trial) and three tails (the 1st, 2nd and 4th trials) = 1/16

Probability of one head (4th trial) and three tails (the 1st, 2nd and 3rd trials) = 1/16

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6 0
3 years ago
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jolli1 [7]

I suppose you mean

g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)

Differentiate one term at a time.

Rewrite the first term as

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Then the product rule says

\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'

Then with the power and chain rules,

\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}

Simplify this a bit by factoring out \frac12 (36-x^2)^{-3/2} :

\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}

For the second term, recall that

\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}

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So we have

g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}

and we can simplify this by factoring out 18(36-x^2)^{-3/2} to end up with

g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}

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Then y=2(4)+8
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