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ss7ja [257]
3 years ago
9

What are the excluded values? 0 and 15 3 and -5 -3 and 5

Mathematics
1 answer:
Dima020 [189]3 years ago
7 0

Answer:

x=5    x = -3

Step-by-step explanation:

-7 / ( x^2 -2x -15)

Factor the denominator

-7

----

(x-5) ( x+3)

The excluded values are when the denominator is equal to zero

(x-5) (x+3) =0

x-5 =0   x+3=0

x=5    x = -3

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Answer the question below
Kobotan [32]

Answer:There are 96 apples in the 12th bag.

Step-by-step explanation:

Firstly, you add the number of apples in the 11 bags.

6+28+16+27+10=87

If the mean number of apples in the 12 bags were 8, the total number of apples are 96.  

Using this info, we subtract 87 from 96 to 9.

Hence, there are 9 apples in the 12th bag.

8 0
3 years ago
Mr.Williams physical education class lasts 7/8 hour if instruciton is 1/5 how many minutes are not spent on instruction?
g100num [7]
= 7/8 − 1/5

= ((7 × 5) − (1 × 8)) / (8 × 5)

= (35 - 8) / 40

= 27/40


SO, 27/40 minutes are NOT spend on instructions.
Hope I was able to help! :)
6 0
3 years ago
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2 years ago
Math help please, thank you!
CaHeK987 [17]
The answer to the first question of the attached document is option 1. We obtain the answer subtracting the term n from the series with the term n-1.For example:
 -3 - (- 5) = 2
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  1 - (- 1) = 2
 So you can see that the common difference is the 2.

 The answer to the second question is option 3:
 y = | x + 7 |
 We can confirm it by substituting values in the equation.
 For example:
 if we do y = 0 then x = -7
 if we do x = 0 then y = 7.
 As corresponds in the graph shown.
 Remember also that as a general rule yes to the equationy = | x | whose vertex is in the point (0,0) we add a positive real number "a" of form y = | x + a | then the graph of y = | x | will move "to" units in the negative direction of x.

 The answer to the third question is option 4.
 The quotient of x and "and" is constant.
 k = y / x
 Rewriting:
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3 0
3 years ago
The plane x+y+2z=8 intersects the paraboloid z=x2+y2 in an ellipse. Find the points on this ellipse that are nearest to and fart
DiKsa [7]

Answer:

The minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

Step-by-step explanation:

Here, the two constraints are

g (x, y, z) = x + y + 2z − 8  

and  

h (x, y, z) = x ² + y² − z.

Any critical  point that we find during the Lagrange multiplier process will satisfy both of these constraints, so we  actually don’t need to find an explicit equation for the ellipse that is their intersection.

Suppose that (x, y, z) is any point that satisfies both of the constraints (and hence is on the ellipse.)

Then the distance from (x, y, z) to the origin is given by

√((x − 0)² + (y − 0)² + (z − 0)² ).

This expression (and its partial derivatives) would be cumbersome to work with, so we will find the the extrema  of the square of the distance. Thus, our objective function is

f(x, y, z) = x ² + y ² + z ²

and

∇f = (2x, 2y, 2z )

λ∇g = (λ, λ, 2λ)

µ∇h = (2µx, 2µy, −µ)

Thus the system we need to solve for (x, y, z) is

                           2x = λ + 2µx                         (1)

                           2y = λ + 2µy                       (2)

                           2z = 2λ − µ                          (3)

                           x + y + 2z = 8                      (4)

                           x ² + y ² − z = 0                     (5)

Subtracting (2) from (1) and factoring gives

                     2 (x − y) = 2µ (x − y)

so µ = 1  whenever x ≠ y. Substituting µ = 1 into (1) gives us λ = 0 and substituting µ = 1 and λ = 0  into (3) gives us  2z = −1  and thus z = − 1 /2 . Subtituting z = − 1 /2  into (4) and (5) gives us

                            x + y − 9 = 0

                         x ² + y ² +  1 /2  = 0

however, x ² + y ² +  1 /2  = 0  has no solution. Thus we must have x = y.

Since we now know x = y, (4) and (5) become

2x + 2z = 8

2x  ² − z = 0

so

z = 4 − x

z = 2x²

Combining these together gives us  2x²  = 4 − x , so

2x²  + x − 4 = 0 which has solutions

x =  (-1+√33)/4

and

x = -(1+√33)/4.

Further substitution yeilds the critical points  

((-1+√33)/4; (-1+√33)/4; (17-√33)/4)   and

(-(1+√33)/4; - (1+√33)/4; (17+√33)/4).

Substituting these into our  objective function gives us

f((-1+√33)/4; (-1+√33)/4; (17-√33)/4) = (195-19√33)/8

f(-(1+√33)/4; - (1+√33)/4; (17+√33)/4) = (195+19√33)/8

Thus minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

4 0
3 years ago
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