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Alexus [3.1K]
3 years ago
5

Write the statement for the problem in mathematical language. Use x for the tens digit and y for the unit digits in the two digi

t numbers. Find the two-digit number which is 2 times the sum of its digits
Mathematics
1 answer:
Nutka1998 [239]3 years ago
8 0

Answer:

  18

Step-by-step explanation:

A two-digit number (xy) will have the value 10x+y. The sum of its digits is x+y.

You want ...

  10x +y = 2(x +y)

  1 ≤ x ≤ 9

  0 ≤ y ≤ 9

The equation can be simplified to ...

  8x = y

The only single-digit values for x and y that satisfy the requirements are ...

  x = 1, y = 8

The two-digit number of interest is 18.

_____

<em>Check</em>

Its sum of digits is 1+8 = 9. 2×9 = 18.

_____

<em>Comment on the relationship in the problem</em>

Many of us learned our multiplication tables for 9s by remembering that the sum of digits of 9n was 9, and that the leading digit of the product was n-1.

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3 years ago
For the differential equations dydx=sqrt(y^2−36) does the existence/uniqueness theorem guarantee that there is a solution to thi
julsineya [31]

Answer:

1. (-4,6) there is no a solution to the equation through this point

2. (2,−6) there is no a solution to the equation through this point

3. (−5,39) there is a solution to the equation through this point

4. (−1,45)  there is a solution to the equation through this point

Step-by-step explanation:

Using the existence and uniqueness theorem:

Let:

F(x,y)=\sqrt{y^2-36} \\\\and\\\\\frac{\partial F}{\partial y} =\frac{y}{\sqrt{y^2-36} }

Now, let's find the domain of F(x,y), due to the square root:

y^2-36 \geq 0\\\\y^2\geq 36\\\\y \geq 6\hspace{12}or\hspace{12}y\leq-6

So the domain of the function is:

y \in R\hspace{12}y\geq6\hspace{12}or\hspace{12}y\leq-6

Now, due to the fraction \frac{\partial F}{\partial y} the denominator must be also different from 0, so:

y^2-36\neq0\\\\y \neq \pm6

So, the theorem  tells us that for each y_0\in R:\hspace{12}y_0>6\hspace{12}or\hspace{12}y_0 there exists a  unique solution defined in an open interval around x_0.

1. (-4,6)  there is no a solution to the equation through this point because y_0=6

2. (2,−6)  there is no a solution to the equation through this point because

y_0=-6

3. (−5,39) there is a solution to the equation through this point because

y_0>6

4. (−1,45)  there is a solution to the equation through this point because

y_0>6

6 0
3 years ago
Find the value of the probability of the standard normal variable Z corresponding to this area for problems 1-3.
Semenov [28]

Answer:

Explained below.

Step-by-step explanation:

The complete question is:

Find the value of the probability of the standard normal variable Z corresponding to this area for problems 1-3.

1. P(Z < 1.62)

2. P(Z > -1.57)

3. P(-1.41 < Z < 0.63)

Solution:

Use Excel to solve the problems.

(1)

P(Z < 1.62)  =NORM.S.DIST(1.62,TRUE)

                 = 0.9474

(2)

P(Z > -1.57)  = P (Z < 1.57)

                  =NORM.S.DIST(1.57,TRUE)

                  = 0.9418

(3)

P(-1.41 < Z < 0.63) = P (Z < 0.63) - P (Z < -1.41)

                             =NORM.S.DIST(0.63,TRUE) - NORM.S.DIST(-1.41,TRUE)

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Inga [223]

Answer:

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Step-by-step explanation:

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This eliminates the first and last inequalities.

Smaller negative numbers are always greater than larger negative numbers (Example: -2 is greater than -4).

This eliminates the third unequality.

For the second given inequality, convert the fractions to share the same denominator.

-2 3/4 < -1 2/4

They are both negative numbers.

We know smaller negative are always greater than larger negative numbers.

This inequality aligns with this statement.

Therefore, only the second inequlity is true.

3 0
3 years ago
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