The table represents a linear function.

Rosa works in an office building with many floors.From her office she goes down 5 floors for a meeting .Then she goes up 8 floor

n200080 [17]

Answer:

11th floor

Step-by-step explanation:

So to figure out what floor Rosa's office is in, we need to declare some variables.

x = Rosa's office

down 5 floors = -5

up 8 floors = 8

down 13 floors = - 13

One major clue to find Rosa's office is that she went down 13 floors to get to the 1st floor. So let's declare that Rosa is on the first floor.

x - 5 + 8 - 13 = 1

x - 10 = 1

x = 1 + 10

x = 11

So Rosa's office is at the 11th floor of the building.

8 0

3 years ago

HELP ME PLEASE ITS NOT THAT HARD

VikaD [51]

Answer:

tan A = 28/45

Step-by-step explanation:

opposite side

the tangent function is defined as tan A = ------------------------

adjacent side

Here the length of the side opposite angle A is 28 and that of the side adjacent to A is 45. Thus, the tangent of angle A is

tan A = 28/45

6 0

4 years ago

What do you need to add 13/4 to make 4

Fudgin [204]

Add 3/4 which becomes 16/4 which becomes 4

3 0

3 years ago

A sample of bacteria is decaying according to a half-life model. If the sample begins with 700 bacteria, and after 10 minutes th

Papessa [141]

Answer:

18 minutes

Step-by-step explanation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is the time,

and T is the half life.

A = 140 when t = 10. Solve for the half life:

140 = 700 (½)^(10 / T)

0.2 = ½^(10 / T)

log 0.2 = (10 / T) log 0.5

10 / T = 2.32

T = 4.31

When A = 40, t is:

40 = 700 (½)^(t / 4.31)

0.057 = ½^(t / 4.31)

log 0.057 = (t / 4.31) log 0.5

t / 4.31 = 4.13

t = 17.8

Rounded to the nearest whole number, it takes 18 minutes.

4 0

4 years ago

Find the absolute maximum and absolute minimum values of the function f(x, y) = x 2 + y 2 − x 2 y + 7 on the set d = {(x, y) : |

dsp73

Looks like $f(x,y)=x^2+y^2-x^2y+7$ .

$f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1$

$f_y=2y-x^2=0\implies2y=x^2$

If $x=0$ , then $y=0$ - critical point at (0, 0).
If $y=1$ , then $x=\pm\sqrt2$ - two critical points at $(-\sqrt2,1)$ and $(\sqrt2,1)$

The latter two critical points occur outside of $D$ since $|\pm\sqrt2|>1$ so we ignore those points.

The Hessian matrix for this function is

$H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}$

The value of its determinant at (0, 0) is $\det H(0,0)=4>0$ , which means a minimum occurs at the point, and we have $f(0,0)=7$ .

Now consider each boundary:

If $x=1$ , then

$f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4$

which has 3 extreme values over the interval $-1\le y\le1$ of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).

If $x=-1$ , then

$f(-1,y)=8-y+y^2$

and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

If $y=1$ , then

$f(x,1)=8$

which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).

If $y=-1$ , then

$f(x,-1)=2x^2+8$

which has 3 extreme values, but the previous cases already include them.

Hence $f(x,y)$ has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

3 0

3 years ago