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3 years ago

15 + (-15) = SOLVEE 10 points

Mathematics

2 answers:

Brrunno [24]3 years ago

6 0

Answer:

The answer is zero 0

Step-by-step explanation:

The problem is saying

15-15

that is 0

prisoha [69]3 years ago

4 0

15+(-15)=0 I hope this helps

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Given f(x) = 2x + 9, if f(x) = 15, find x.

AleksandrR [38]

Answer:

Step-by-step explanation:

$f(x)=2x+9\\\\\text{Substitute}\ f(x)=15:\\\\15=2x+9\qquad\text{subtract 9 from both isdes}\\\\15-9=2x+9-9\\\\6=2x\qquad\text{divide both sides by 2}\\\\\dfrac{6}{2}=\dfrac{2x}{2}\\\\3=x\to x=3$

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3 years ago

36^x-2=6 solve for x

Lina20 [59]

Answer:

x = $\frac{in (8)}{in (36)}$ if you consider doing decimal form x = 0.58027921. . .

Step-by-step explanation:

Take logarithm of both sides of the equation to remove the variable from the exponent.

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2 years ago

Which real-world story is represented by the addition problem (–2) + (–3) = –5?

tangare [24]

The answer is A.

If the submarine descended then it went below, and if it ascended it went above. Therefore:

"Descended 2 miles"= -2

"Then descended another 3 miles"= -3

(-2)+(-3)=-5

Adding two negatives together will give you a bigger negative so the submarine would then be 5 miles below its original location.

6 0

3 years ago

How would you convert 276 yards to inches?

valentinak56 [21]

Answer:

9936 inch.

Step-by-step explanation:

Note the measurements:

1 Yard = 3 Feet

1 Feet = 12 Inch

1 Yard = 3 Feet = (12)(3) Inches = 36 Inches; 1 Yard = 36 Inches

Multiply 276 with 36

276 x 36 = 9936

9936 inches is your answer.

6 0

3 years ago

The plane x + y + z = 12 intersects paraboloid z = x^2 + y^2 in an ellipse.(a) Find the highest and the lowest points on the ell

emmasim [6.3K]

Answer:

Highest (-3,-3)

Lowest (2,2)

Farthest (-3,-3)

Closest (2,2)

Step-by-step explanation:

To solve this problem we will be using Lagrange multipliers.

Let us find out first the restriction, which is the projection of the intersection on the XY-plane.

From x+y+z=12 we get z=12-x-y and replace this in the equation of the paraboloid:

$\bf 12-x-y=x^2+y^2\Rightarrow x^2+y^2+x+y=12$

completing the squares:

$\bf x^2+y^2+x+y=12\Rightarrow (x+1/2)^2-1/4+(y+1/2)^2-1/4=12\Rightarrow\\\\\Rightarrow (x+1/2)^2+(y+1/2)^2=12+1/2\Rightarrow (x+1/2)^2+(y+1/2)^2=25/2$

and we want the maximum and minimum of the paraboloid when (x,y) varies on the circumference we just found. That is, we want the maximum and minimum of

$\bf f(x,y)=x^2+y^2$

subject to the constraint

$\bf g(x,y)=(x+1/2)^2+(y+1/2)^2-25/2=0$

Now we have

$\bf \nabla f=(\displaystyle\frac{\partial f}{\partial x},\displaystyle\frac{\partial f}{\partial y})=(2x,2y)\\\\\nabla g=(\displaystyle\frac{\partial g}{\partial x},\displaystyle\frac{\partial g}{\partial y})=(2x+1,2y+1)$

Let $\bf \lambda$ be the Lagrange multiplier.

The maximum and minimum must occur at points where

$\bf \nabla f=\lambda\nabla g$

that is,

$\bf (2x,2y)=\lambda(2x+1,2y+1)\Rightarrow 2x=\lambda (2x+1)\;,2y=\lambda (2y+1)$

we can assume (x,y)≠ (-1/2, -1/2) since that point is not in the restriction, so

$\bf \lambda=\displaystyle\frac{2x}{(2x+1)} \;,\lambda=\displaystyle\frac{2y}{(2y+1)}\Rightarrow \displaystyle\frac{2x}{(2x+1)}=\displaystyle\frac{2y}{(2y+1)}\Rightarrow\\\\\Rightarrow 2x(2y+1)=2y(2x+1)\Rightarrow 4xy+2x=4xy+2y\Rightarrow\\\\\Rightarrow x=y$

Replacing in the constraint

$\bf (x+1/2)^2+(x+1/2)^2-25/2=0\Rightarrow (x+1/2)^2=25/4\Rightarrow\\\\\Rightarrow |x+1/2|=5/2$

from this we get

and the candidates for maximum and minimum are (2,2) and (-3,-3).

Replacing these values in f, we see that

f(-3,-3) = 9+9 = 18 is the maximum and

f(2,2) = 4+4 = 8 is the minimum

Since the square of the distance from any given point (x,y) on the paraboloid to (0,0) is f(x,y) itself, the maximum and minimum of the distance are reached at the points we just found.

We have then,

(-3,-3) is the farthest from the origin

(2,2) is the closest to the origin.

3 0

3 years ago