–12 – 4.8 ∙ (–2 + –8) whats this equal need hlp asap 10 pts and i will befreind

Which of the following expressions is this one equivalent to?

schepotkina [342]

Answer: C. x-2

Step-by-step explanation:

We have the following expression:

$\frac{x^{4} -2x^{3} +x-2}{x^{3}+1}$

Factoring in the numerator:

$\frac{x(x^{3}+1)-2(x^{3}+1)}{x^{3}+1}$

Factoring again in the numerator with common factor $x^{3}+1$ :

$\frac{(x^{3}+1)(x-2)}{x^{3}+1}$

Simplifying:

$x-2$

Hence, the correct option is C. x-2

3 0

3 years ago

57 students choose to attend one of three after school activities: football, tennis or running.

skelet666 [1.2K]

The probability that the randomly selected student chose running is; 12/57

<h3>Solving Probability Questions</h3>

Total number of students = 57

Total number of boys = 24

Total number of girls = 33

Number of Girls that chose football = 17

Number of boys that chose football = 14

Number of students that chose Tennis = 14

Now, number of students that chose running will be;

Number of students that chose running = 57 - (31 + 14) = 12

Thus, probability that a randomly selected student chose running = 12/57

Read more about probability selection at; brainly.com/question/251701

3 0

3 years ago

Find two vectors in R2 with Euclidian Norm 1<br> whoseEuclidian inner product with (3,1) is zero.

alina1380 [7]

Answer:

$v_1=(\frac{1}{10},-\frac{3}{10})$

$v_2=(-\frac{1}{10},\frac{3}{10})$

Step-by-step explanation:

First we define two generic vectors in our $\mathbb{R}^2$ space:

$v_1 = (x_1,y_1)$
$v_2 = (x_2,y_2)$

By definition we know that Euclidean norm on an 2-dimensional Euclidean space $\mathbb{R}^2$ is:

$\left \| v \right \|= \sqrt{x^2+y^2}$

Also we know that the inner product in $\mathbb{R}^2$ space is defined as:

$v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2$

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

$\left \| v_1 \right \|= \sqrt{x^2+y^2}=1$

and

$\left \| v_2 \right \|= \sqrt{x^2+y^2}=1$

As second condition we have that:

$v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0$

$v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0$

Which is the same:

$y_1=-3x_1\\y_2=-3x_2$

Replacing the second condition on the first condition we have:

$\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}$

Since $x_1^2= \frac{1}{10}$ we have two posible solutions, $x_1=\frac{1}{10}$ or $x_1=-\frac{1}{10}$ . If we choose $x_1=\frac{1}{10}$ , we can choose next the other solution for $x_2$ .

Remembering,

$y_1=-3x_1\\y_2=-3x_2$

The two vectors we are looking for are:

$v_1=(\frac{1}{10},-\frac{3}{10})\\v_2=(-\frac{1}{10},\frac{3}{10})$

5 0

3 years ago

A gravel path is to be built around an 18‘ x 15‘ rectangle garden if the width remains constant how why can the path be if there

Ilia_Sergeevich [38]

Answer:

P = 2*(18 + 13) = 62ft

Now to figure out the width of the path we have to do some geometry. There are 3 basic shapes we will use for the path, they are the 13 ft long rectangle, the 18 ft long rectangle, and the squares formed at the corners. These squares' areas are equal to the width squared, whereas the rectangles are only equal to the length times the width. Our total area then becomes:

A = 2*(13*w) + 2*(18*w) + 4*(w*w)

A = 4w^2 + 62w

And since we know the available area is 516 ft^2, putting this in for A and solving the quadratic equation we can get the width:

516 = 4w^2 + 62w

0 = 4w^2 + 62w - 516

Using the quadratic formula: (w = -b2+/-SQRT(b^2 - 4ac)/2a, where a = 4, b = 62, and c = -516)

w = 6, -21.5

Since we obviously cannot have a negative width, 6 must be our answer. Therefore the path can be 6 feet wide!

We can check this answer by doing the following. Please note that by having a 6ft wide path you would actually be adding 12 ft to both dimensions, since you add 6 feet of length to both sides. Your total area would then become:

18+(2*6) x 13+(2*6) = 30x25 = 750 ft^2

Subtracting the area of the garden gives you the are of the gravel required:

750ft^2 - (18*13)ft^2 = 516ft^2

Therefore a width of 6 feet is correct.

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

I need help on 33 please help

lyudmila [28]

Uhh I think it a carrot

7 0

3 years ago