<img src="https://tex.z-dn.net/?f=%7Cx-1%7C%3D%5Csqrt%7Bx%7D" id="TexFormula1" title="|x-1|=\sqrt{x}" alt="|x-1|=\sqrt{x}" align

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kenny6666 [7]

3 years ago

="absmiddle" class="latex-formula">

Mathematics

1 answer:

vodka [1.7K]3 years ago

7 0

Isolate the radical, then raise each side of the equation to the power of its index.
Exact Form:
x
=
3
+
√
5
2
,
3
−
√
5
2
x
=
3
+
5
2
,
3
-
5
2

Decimal Form:
x
=
2.61803398
…
,
0.38196601
…

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joan is rolling a six— sided numbers cube with 1—6 on the sides. what is the probability that she will roll a numbers less tha 3

Ulleksa [173]

1/3 because only 1 and 2 is less than 3 out of the 6 numbers. 2/6 = 1/3.

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3 years ago

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Slope intercept form of y-3=2/3(x+5)

Blizzard [7]

Answer:

It's already in slope intercept form

slope = 2/3

Intercept = (-5,3)

Step-by-step explanation:

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2 years ago

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Travka [436]

Answer:

see below

Step-by-step explanation:

y= 1/2x - 2

To find the x intercept, set y =0 and solve for x

0= 1/2x - 2

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3 years ago

Does the graph represent a function ? Why or Why not ?

Bogdan [553]

Answer:yes

Step-by-step explanation: the answer does represent a function because if you were to put points in the line wherever, and connect them by drawing a line vertically, it would not cross two points.

3 0

3 years ago

WHAT IS THE REMAINDER WHEN <img src="https://tex.z-dn.net/?f=32%5E%7B37%5E%7B32%7D%20%7D" id="TexFormula1" title="32^{37^{32} }"

Feliz [49]

Recall Euler's theorem: if $\gcd(a,n) = 1$ , then

$a^{\phi(n)} \equiv 1 \pmod n$

where $\phi$ is Euler's totient function.

We have $\gcd(9,32) = 1$ - in fact, $\gcd(9,32^k)=1$ for any $k\in\Bbb N$ since $9=3^2$ and $32=2^5$ share no common divisors - as well as $\phi(9) = 6$ .

Now,

$37^{32} = (1 + 36)^{32} \\\\ ~~~~~~~~ = 1 + 36c_1 + 36^2c_2 + 36^3c_3+\cdots+36^{32}c_{32} \\\\ ~~~~~~~~ = 1 + 6 \left(6c_1 + 6^3c_2 + 6^5c_3 + \cdots + 6^{63}c_{32}\right) \\\\ \implies 32^{37^{32}} = 32^{1 + 6(\cdots)} = 32\cdot\left(32^{(\cdots)}\right)^6$

where the $c_i$ are positive integer coefficients from the binomial expansion. By Euler's theorem,

$\left(32^{(\cdots)\right)^6 \equiv 1 \pmod9$

so that

$32^{37^{32}} \equiv 32\cdot1 \equiv \boxed{5} \pmod9$

7 0

1 year ago