When 142 is added to a number, the result is 64 more than 3 times the number. Find

1 answer:

8 0

X+142 =3x+64

subtract x from each side

142 = 2x+64

subtract 64 from each side

142-64 = 2x

78=2x

divide by 2

78/2 = x

x=39

The number is 39

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Find the remainder when f(x)=x3−4x2−6x−3 f ( x ) = x 3 − 4 x 2 − 6 x − 3 is divided by x+1

loris [4]

Answer:

The remainder is -2.

Step-by-step explanation:

According to the Polynomial Remainder Theorem, if we divide a polynomial P(x) by a binomial (x - a), then the remainder of the operation will be given by P(a).

Our polynomial is:

$P(x) = x^3-4x^2-6x-3$

And we want to find the remainder when it's divided by the binomial:

$x+1$

We can rewrite our divisor as (x - (-1)). Hence, a = -1.

Then by the PRT, the remainder will be:

$\displaystyle\begin{aligned} R &= P(-1)\\ &=(-1)^3-4(-1)^2-6(-1)-3 \\ &= (-1)-4(1)+(6)-3 \\ &= -2 \end{aligned}$

The remainder is -2.

5 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,

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