Ask question

Ask question

All categories

bearhunter [10]

3 years ago

11

Please help with 25 I need this

2 answers:

notka56 [123]3 years ago

8 0

Point A is (2,1)
Point B is (5,2)
Point C is (3,5)

tino4ka555 [31]3 years ago

3 0

$48 \: ft {}^{2}$
It would be 8 feet long by and 6 feet wide

You might be interested in

CAN SOMEONE PLEASE HELP ME WITH THESE PEOPLE

vazorg [7]

Inverse of $f(x)=x+2$ is $f^{-1}(x)=x-2$ because

$f(x)=x+2\Rightarrow x=f^{-1}(x)+2\Rightarrow\boxed{f^{-1}(x)=x-2}$

Now $f^{-1}(-5)=-5-2=\boxed{-7}$ .

Hope this helps.

4 0

3 years ago

Which of the below statements is correct and then explain your answer

VMariaS [17]

Answer:

Greater than 3

Step-by-step explanation:

Logarithms explain the relationship between exponents.

Think of them as such, except they got their bases and exponents switched.

In this logarithm:

$log2(10)$

basically says 2 to the power of <em>what</em> gets you to 10.

Now let's experiment: 2^2=4, 2^3=8. 2^4=16.

Wait, we already crossed 10. Now we know that log2(10) must be in between 3 and 4. We don't need to find the exact value, because thats not what the question asked.

It's between 3 and 4, so it has to be greater than 3 but less than 4. Thee is your answer!

Have a nice day! :)

8 0

2 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

What is the volume of a Sphere with the radius of 12cm?

kramer

Answer:

I think the answer would be

V≈7238.23cm³

6 0

3 years ago

Read 2 more answers

How many ways to make a password with 2 letters followed by 3 numbers if NO<br> repeats are allowed?

faust18 [17]

Answer:

my105

Step-by-step explanation:

I don't really know if this will work but

Hope it helped: )

5 0

3 years ago

Read 2 more answers