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MA_775_DIABLO [31]

3 years ago

11

The cube of the difference of 1 and a number

1 answer:

skelet666 [1.2K]3 years ago

8 0

Answer:

$(1-x)^3$

Step-by-step explanation:

Let

x ----> the number

we know that

The cube of the difference of 1 and a number is the same that subtract the number from 1 and all the expression elevated to the cube

so

$(1-x)^3$

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Whats -5 5/8 divided by -4 7/12 as a fraction?

sesenic [268]

Turn the mixed numbers into improper fractions:
-5 5/8 = - 45/8
-4 7/12 = - 55/12

Remember, when you divide by a fraction, you take its reciprocal and multiply it:
(-48/8) / (-55/12) = (-48/8) x (-12/55)

Simplify the fraction:
72/55

I think this is the answer.

8 0

3 years ago

What is the y intercept of the function

Dahasolnce [82]

Answer:

Do you have a picture you could attach to it? That would greatly help! :)

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

If you were to roll the dice one time what is the probability of it landing on an odd number?

yanalaym [24]

Answer:

1/2

Step-by-step explanation:

a dice has six sides and the highest number is 6

the dice should have the numbers

1,2,3,4,5,6

which 1,3,5 are odd numbers so it's a 1/2

3 0

3 years ago

Plot, compare, and order √27, √25, √60+ 2² -2.

raketka [301]

The numbers in increasing order are ∛27, √25 and √(60+ 2²) -2

<h3>How to plot, compare, and order the numbers?</h3>

The numbers are given as

∛27, √25, √(60+ 2²) -2

Evaluate the expressions

So, we have:

∛27 = 3

√25 = 5

√(60+ 2²) -2 = 6

Next, we order the numbers in ascending order.

So, we have:

∛27 = 3

√25 = 5

√(60+ 2²) -2 = 6

Remove the evaluated values

So, we have:

∛27, √25 √(60+ 2²) -2

Next, we plot the numbers on a number line

Read more about number lines at:

brainly.com/question/24644930

#SPJ1

5 0

2 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