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Lina20 [59]
3 years ago
14

Find the remainder when 51^203 is divided by 7 (using remainder theorem).

Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
6 0
Remainder Theorem is an application of Euclidean Division of polynomials and states that the remainder of the division of a polynomial f(x) by a linear polynomial x-1 is equal to f(a) so based on your question the answer would be the remainder is 4.
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HELP PLS AND BE QUICK!!! I NEED IT SOON
enyata [817]

Answer:

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Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Evaluate-2+7x^2-2x^3-8x+7x^4 at x=5
Murljashka [212]

Answer:

4262

Step-by-step explanation:

2 + 7x^2- 2x^3-8x +7x^4     at x = 5

= 2 + 7*(5)^2 - 2*(5)^3-8*5 +7*(5)^4

=2+175-250-40+4375

=4552-290

=4262

4 0
3 years ago
Which of the following functions is graphed below.
Valentin [98]

Answer: Option A

y=\left \{ {{x^2 +2;\ \ x

Step-by-step explanation:

In the graph we have a piecewise function composed of a parabola and a line.

The parabola has the vertex in the point (0, 2) and cuts the y-axis in y = 2.

The equation of this parabola is y = x ^ 2 +2

Then we have an equation liney = -x + 2


Note that the interval in which the parabola is defined is from -∞ to x = 1. Note that the parabola does not include the point x = 1 because it is marked with an empty circle " о ."

(this is x< 1)

Then the equation of the line goes from x = 1 to ∞ . In this case, the line includes x = 1 because the point at the end of the line is represented by a full circle .

(this is x\geq 1)

Then the function is:

y=\left \{ {{x^2 +2;\ \ x

7 0
3 years ago
Considering only the values of β for which sinβtanβsecβcotβ is defined, which of the following expressions is equivalent to sinβ
-Dominant- [34]

Answer:

\tan(\beta)

Step-by-step explanation:

For many of these identities, it is helpful to convert everything to sine and cosine, see what cancels, and then work to build out to something.  If you have options that you're building toward, aim toward one of them.

{\tan(\theta)}={\dfrac{\sin(\theta)}{\cos(\theta)}    and   {\sec(\theta)}={\dfrac{1}{\cos(\theta)}

Recall the following reciprocal identity:

\cot(\theta)=\dfrac{1}{\tan(\theta)}=\dfrac{1}{ \left ( \dfrac{\sin(\theta)}{\cos(\theta)} \right )} =\dfrac{\cos(\theta)}{\sin(\theta)}

So, the original expression can be written in terms of only sines and cosines:

\sin(\beta)\tan(\beta)\sec(\beta)\cot(\beta)

\sin(\beta) * \dfrac{\sin(\beta) }{\cos(\beta) } * \dfrac{1 }{\cos(\beta) } * \dfrac{\cos(\beta) } {\sin(\beta) }

\sin(\beta) * \dfrac{\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} * \dfrac{1 }{\cos(\beta) } * \dfrac{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} {\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}

\sin(\beta) *\dfrac{1 }{\cos(\beta) }

\dfrac{\sin(\beta)}{\cos(\beta) }

Working toward one of the answers provided, this is the tangent function.


The one caveat is that the original expression also was undefined for values of beta that caused the sine function to be zero, whereas this simplified function is only undefined for values of beta where the cosine is equal to zero.  However, the questions states that we are only considering values for which the original expression is defined, so, excluding those values of beta, the original expression is equivalent to \tan(\beta).

8 0
2 years ago
What is the relationship between variance and standard deviation?
vovangra [49]
The relationship between variance and standard deviation is :
Standard deviation = sqrt (variance),
Variance = standard deviation^2.

hope this help
8 0
3 years ago
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