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Dafna11 [192]
3 years ago
11

Help Needed, Much Thanks Generous amount of points4a+bx^Find X

Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
7 0

I believe x would just be 4? There isn’t much information given so that’s my best guess.

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HELP!! What is the approximate measure of angle x in the triangle shown?
Alex Ar [27]
<h3>Answer:   D)  130.5 degrees</h3>

=================================================

Work Shown:

c^2 = a^2 + b^2 - 2*a*b*\cos(C)\\\\10^2 = 5^2 + 6^2 - 2*5*6*\cos(x)\\\\100 = 25 + 36 - 60*\cos(x)\\\\100 = 61 - 60*\cos(x)\\\\100 - 61 = - 60*\cos(x)\\\\39 = - 60*\cos(x)\\\\\cos(x) = \frac{39}{-60}\\\\\cos(x) = -0.65\\\\x = \arccos(-0.65)\\\\x \approx 130.5416\\\\x \approx 130.5\\\\

Note: I used the law of cosines. Make sure your calculator is in degree mode.

6 0
2 years ago
Please help me ASAP
Liula [17]
I would say answer choice 1. It’s the only one that makes sense.

I hope this helps! Let me know if you have any questions! Good luck on the rest of your assessment!

Mark this as Brainliest if you get a chance:)
6 0
2 years ago
Read 2 more answers
John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m
Dmitry [639]

Answer:

The minimum possible initial amount of fish:52

Step-by-step explanation:

Let's start by saying that

x = is the initial number of fishes

John:

When John arrives:

  • he throws away one fish from the bunch

x-1

  • divides the remaining fish into three.

\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}

  • takes a third for himself.

\dfrac{x-1}{3} + \dfrac{x-1}{3}

the remaining fish are expressed by the above expression. Let's call it John

\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}

and simplify it!

\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}

When Joe arrives:

  • he throws away one fish from the remaining bunch

\text{John} -1

  • divides the remaining fish into three

\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}

  • takes a third for himself.

\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

the remaining fish are expressed by the above expression. Let's call it Joe

\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

and simiplify it

\text{Joe}=\dfrac{2}{3}(\text{John}-1)

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)

\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}

When James arrives:

We're gonna do this one quickly, since its the same process all over again

\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}

\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)

\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say n), a third is taken away by James. the remaining bunch would be \frac{n}{3}+\frac{n}{3}

hence we've expressed the last pile in terms of n as well.  Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

14=\dfrac{8x}{27} - \dfrac{38}{27}

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

14=\dfrac{8x}{27} - \dfrac{38}{27}

x=52

This is minimum possible amount of fish before John threw out the first fish

8 0
2 years ago
S(t)=sqrt3t+1 at t=2
avanturin [10]

Answer:

s(2) = √7

Step-by-step explanation:

s(2) = √(3[2] + 1) = √7

5 0
2 years ago
A rectangular rug is 12.5 feet long and 10 feet wide. The rug needs to be reduced by a factor of One-fifth to be placed near a d
11111nata11111 [884]
You would do 12.5 x10 then divided it by 1/5
3 0
2 years ago
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