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3 years ago

One more than 8 times a number is the same as 12 times the number

1 answer:

Nadusha1986 [10]3 years ago

7 0

Using algebra makes stuff like this easy.
8x + 1 = 12x - 3
4 = 4x
The number is 1!
One more than 8 times 1 is 9.
12 times one, decreased by 3, is also 9.

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Solve 6 over x minus 3 equals 3 over x for x and determine if the solution is extraneous or not

Natali5045456 [20]

Answer:

This is the equation that I interpreted:

6/(x) - 3 = 3/(x)

I have never worked with determining extraneous equations before, so I cannot answer the second part.

8 0

3 years ago

What is the quotient of 5 divided by 1/3

Pachacha [2.7K]

Answer:

Step-by-step explanation:

5 divided by 1/3 is the same as if you flip the denominator meaning 5 x 3 which is 15

4 0

3 years ago

Read 2 more answers

F(x) = x2 + 4x − 6

german

Since

$x^2+4x+4=(x+2)^2$

Is a perfect square, we can think of the "-6" at the end as a "+4-10" and we have

$x^2+4x-6 = x^2+4x+4-10 = (x+2)^2-10$

Which is the required form

6 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

Either enter an exact answer in terms of \piπpi or use 3.14, point, 14 for \piπpi and enter your answer as a decimal.

OverLord2011 [107]

Answer:

49π sq. units

Step-by-step explanation:

Find the diagram attached

We can find the area of the given circle

Given

diameter of the circle d = 14

Area of the circle = πd²/4

Substitute the given diameter

Area of the circle = π(14)²/4

Area of the circle = 196π/4

Area of the circle = 49π sq. units

Hence the exact answer in terms of π is 49π sq. units

4 0

3 years ago