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brilliants [131]
2 years ago
15

−1 − (−z) where z= -2

Mathematics
2 answers:
Tpy6a [65]2 years ago
6 0

Answer:

-3

Step-by-step explanation:

Stels [109]2 years ago
5 0

Answer: -3

Step-by-step explanation:

-1 - (--2)

-1 - 2

= -3

Whenever you see 2 negative , or - signs next to each other, like --, they cancel out.

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a salesperson earns $25 per day plus $2.50 for every sale. He normally earns $100 per day. How many sales does he normally get.
Dahasolnce [82]
He makes about 30 sales because he makes $100 typically but always makes a minimum of $25 making $75 worth of his sales. If you multiply 30 and $2.50 you get $75 in sales. His $25 minimum and $75 in sales make $100. So he makes 30 sales.
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What is 10x + 5 = -5
BabaBlast [244]

Answer:

x = -1

Step-by-step explanation:

10x + 5 = -5

      - 5    -5

10x = -10

10 / -10 = -1

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Step-by-step explanation

7 0
3 years ago
Consider the function f (x comma y )equals3 x squared minus 2 y squared minus 3 and the point (negative 2 comma 1 ). a. Find the
Dima020 [189]

Answer: see description

Step-by-step explanation:

given the function f(x,y) = 3x^2-2y^2-3

we calculate the gradient \nabla f(x,y) = \frac{\partial f(x,y)}{\partial x} \hat{\textbf{x}}+ \frac{\partial f(x,y)}{\partial y} \hat{\textbf{y}} + \frac{\partial f(x,y)}{\partial z} \hat{\textbf{z}}

for each term we consider all variables different to the one we are derivating as constants. For each term we have

\frac{\partial f(x,y)}{\partial x} = 3 \frac{\partial }{\partial x}(x^2) = 6x\\ \frac{\partial f(x,y)}{\partial y} = -2 \frac{\partial }{\partial x}(y^2) = -4y\\\frac{\partial f(x,y)}{\partial z} = 0\\

Therefore:

\nabla f(x,y) = 6x \hat{\textbf{x}} -4y \hat{\textbf{y}}

gives the direction of maxium increase.

a) with x = -2, y= 1

\nabla f(-2,1) = which magnitude is \sqrt{(-12)^2+(-4)^2 } = \sqrt{160} = 4 \sqrt{10}

so the unitary vector in the direction of the steepest ascent is  

u_{1} = \frac{\nabla f(-2,1)}{|\nabla f(-2,1)|} = \frac{1}{4 \sqrt(10)}*

and the unitary vector in the direction of steepest descent is

u_{2} = \frac{- \nabla f(-2,1)}{|\nabla f(-2,1)|} = \frac{1}{4 \sqrt(10)}*

finally, the vector in no change direction is basically doing one of the following possibilities with u_{1}:

if we have a vector <a,b> the perpendicular vector (direction of no change) will be either <-a,b> or <a,-b>

so i will select <-a,b>

u_{no change} =  \frac{1}{4 \sqrt(10)}*

4 0
3 years ago
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