Graphing a line it’s equation in standard form

What is the sum of the squares of 7 and 24 ?

qwelly [4]

Answer:

625

Step-by-step explanation:

8 0

3 years ago

Which could be used to evaluate the expression Negative 6 (4 and two-thirds)?

Firdavs [7]

Answer:

The answer is (Negative 6) (4) + (negative 6) (two-thirds)

Step-by-step explanation:

So basically, Negative 6x4= negative 24. Then you have to add Negative 6 x two-thirds. So 6 times two-thirds equal 4. So now you just add 24 and 4. And that equals 28. :)

6 0

3 years ago

Read 2 more answers

Which step is included in the graph of the function f(x)=[x-1]? (the brackets are ceiling functions symbols)

KiRa [710]

we are given

f(x)=[x=1]

where bracket means ceiling functions

we know that

Ceiling function returns the least value of the integer that is greater than or equal to the specified number

so, we can check each options

option-A:

$-4\leq x$

At x=-4:

f(x)=[-4-1] =-5

For x<-3:

Let's assume

x=-3.1

f(x)=[-3.1-1] =[-4.1]=-5

so, this interval is TRUE

option-B:

$-2\leq x$

At x=-2:

f(x)=[-2-1] =-3

For x<-1:

Let's assume

x=-1.1

f(x)=[-1.1-1] =[-2.1]=-3

so, this is FALSE

7 0

4 years ago

Read 2 more answers

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 $x_2$ .