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ira [324]
2 years ago
10

Determine the smallest integer value of x in the solution 4x-9 ≥ 17

Mathematics
2 answers:
Ghella [55]2 years ago
6 0

Answer:

7 is the smallest integer value

plz brainliestt

Step-by-step explanation:

4x>_ 26

x>_6.5

jeka942 years ago
3 0

Answer:

6.5

Step-by-step explanation:

calculator

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A basketball player makes 17 of the 25 shots she takes. What percentage of shots did she successfully make?
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\pi {4}^{2}  -  \: \pi {2}^{2}

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3 years ago
Determine whether the following sequence is geometric. If so, find the common ratio.
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8 0
3 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
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