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a_sh-v [17]
2 years ago
6

How many different four-digit numbers can be formed with the numbers 7; 4; 5; 1; 2; 9; and 8?

Mathematics
1 answer:
Masja [62]2 years ago
7 0

Answer:

If I'm correct, the answer is 1296.

Step-by-step explanation:

6 x 6 x 6 x 6 = 1296

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Find an equation of the tangent to the curve x =5+lnt, y=t2+5 at the point (5,6) by both eliminating the parameter and without e
svet-max [94.6K]

ANSWER

y = 2x -4

EXPLANATION

Part a)

Eliminating the parameter:

The parametric equation is

x = 5 +  ln(t)

y =  {t}^{2}  + 5

From the first equation we make t the subject to get;

x - 5 =  ln(t)

t =  {e}^{x - 5}

We put it into the second equation.

y =  { ({e}^{x - 5}) }^{2}  + 5

y =  { ({e}^{2(x - 5)}) }  + 5

We differentiate to get;

\frac{dy}{dx}  = 2 {e}^{2(x - 5)}

At x=5,

\frac{dy}{dx}  = 2 {e}^{2(5 - 5)}

\frac{dy}{dx}  = 2 {e}^{0}  = 2

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

y-y_1=m(x-x_1)

y - 6 = 2(x - 5)

y = 2x - 10 + 6

y = 2x -4

Without eliminating the parameter,

\frac{dy}{dx}  =  \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }

\frac{dy}{dx}  =  \frac{ 2t}{  \frac{1}{t} }

\frac{dy}{dx}  =  2 {t}^{2}

At x=5,

5 = 5 +  ln(t)

ln(t)  = 0

t =  {e}^{0}  = 1

This implies that,

\frac{dy}{dx}  =  2 {(1)}^{2}  = 2

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

y-y_1=m(x-x_1)

y - 6 = 2(x - 5) =

y = 2x -4

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3 years ago
Let g be the function given by g(x)=√ 1−sin2x. Which of the following statements could be false on the interval 0≤x≤π ?
zysi [14]

Answer:

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Step-by-step explanation:

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6 0
2 years ago
Solve the system of equations by finding the reduced row echelon form of the augmented matrix. Write the solutions for x and y i
Gwar [14]

Answer:

The solutions for the given system of equations are:

\left \{ {{x=-5} \atop {y=12-4z}} \right.

Step-by-step explanation:

Given the equation system:

\left \{ {{3x+y+4z=-3} \atop {-x+y+4z=17}} \right.

We obtain the following matrix:

\left[\begin{array}{cccc}3&1&4&-3\\-1&1&4&17\end{array}\right]

<u>Step 1:</u> Multiply the fisrt row by 1/3.

\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\-1&1&4&17\end{array}\right]

<u>Step 2:</u> Sum the first row and the second row.

\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\0&\frac{4}{3} &\frac{16}{3}&16\end{array}\right]

<u>Step 3:</u> Multiply the second row by 3/4.

\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\0&1 &4&12\end{array}\right]

<u>Step 4:</u> Multiply the second row by -1/3 and sum the the first row.

\left[\begin{array}{cccc}1&0 &0&-5\\0&1 &4&12\end{array}\right]

The result of the reduced matrix is:

\left \{ {{x=-5} \atop {y+4z=12}} \right.

This is equal to:

\left \{ {{x=-5} \atop {y=12-4z}} \right.

These are the solutions for the system of equations in terms of z, where z can be any number.

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In circle c with diameter SP, what is the measure of PCQ?
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Answer:

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