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2 years ago

HELPPPPPPPPPPPPPPPPPPPPPPPPPP

Mathematics

1 answer:

beks73 [17]2 years ago

3 0

Answer:

c 150

Step-by-step explanation:

because opposite angle have same angles

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A number cube is rolled 360 times and the results are recorded as follows: 47 ones, 55 twos, 67 threes, 67 fours, 35 fives, and

Ede4ka [16]

122/360 which can be simplified to 61/180 which 34% chance.

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2 years ago

5x^(-2)-17x^(-1)+6<br> i’m so confused pls help<br> will give brainliest

laila [671]

It's a quadratic equation in disguise. If you let $y=x^{-1}$ , then $y^2=x^{-2}$ , and we can rewrite the equation as

$5y^2-17y+6=0$

Solve for $y$ however you like; we get $y=\dfrac25$ and $y=3$ .

But we want to solve for $x$ , so we have

$y=x^{-1}=\dfrac25\implies x=\dfrac52$

$y=x^{-1}=3\implies x=\dfrac13$

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3 years ago

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jenyasd209 [6]

The expression that represents the number of days until only 10% remains is

T((d) 10%) = 100 • (1/2)^3.322

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3 years ago

HELP ME PLZ

GarryVolchara [31]

Answer:

(-1,3)

Step-by-step explanation:

Thus;

When (-1,3) is placed in the relation 3x + y =0,the answer will be zero.

<h3>Calculations</h3>

Let;

x= -1

y= 3

3(-1) + 3 =0

-3 + 3 = 0

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2 years ago

Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N defini

Nat2105 [25]

By inspection, it's clear that the sequence must converge to $\dfrac32$ because

$\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32$

when $n$ is arbitrarily large.

Now, for the limit as $n\to\infty$ to be equal to $\dfrac32$ is to say that for any $\varepsilon>0$ , there exists some $N$ such that whenever $n>N$ , it follows that

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|$

From this inequality, we get

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}$
$\implies|2n-1|>\dfrac5{2\varepsilon}$
$\implies2n-1\dfrac5{2\varepsilon}$
$\implies n\dfrac12+\dfrac5{4\varepsilon}$

As we're considering $n\to\infty$ , we can omit the first inequality.

We can then see that choosing $N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil$ will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that $N\in\mathbb N$ .

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3 years ago