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3 years ago

What is the value of x if gx-1 -2 =25? x=1/2 x=2 x=5/2 x=4

Mathematics

1 answer:

Julli [10]3 years ago

5 0

Answer:

x=5/2

Step-by-step explanation:

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6+0.4+0.009___6+0.07+0.009 which statement compares the two numbers correctly

VikaD [51]

Answer:

Step-by-step explanation:

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3 years ago

RSM QUESTION NEED HELP!!!!!! A pharmacist has a 13 % alcohol solution and another 18 % alcohol solution. How much of each must h

Luden [163]

Answer:

40 grams of 13% alcohol solution and 10 grams of 18% alcohol solution.

Step-by-step explanation:

13%(x) + 18%(50-x) = 14%*50g Multiply them all

0.13x + (9-0.18x) = 7

0.13x-0.18x + 9 = 7 Simplify the equation and subtract 0.13x from 0.18x

9-0.05x = 7 Add 0.05x to each side

9 = 7+0.05x Subtract 7 from both sides

0.05x = 2 Multiply each side by 100

5x = 200 Divide both sides by 5

x = 40

50 - 40 = 10

40 grams and 10 grams

Your welcome

4 0

3 years ago

Read 2 more answers

A kitchen makes 2,944 ounces of soup each week. There are 128 ounces in 1 gallon. How many gallons of soup does the kitchen make

Rashid [163]

I calculated its 14 gallons is in two week
1 gall=128
128 x 1st week= 896
128 x 2nd week=1792divid it by 128=14
mark me as brainlest

8 0

3 years ago

Solve the system of equations below by graphing.

AleksandrR [38]

I think it would be B hope it helps

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3 years ago

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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