Who wants to be my friend? :)

PLEASE HELP IM ON A TIME LIMIT :(

Mice21 [21]

Answer:

hey hope this helps

<h3 /><h3>Comparing sides AB and DE </h3>

AB =

$\sqrt{ {1}^{2} + {1}^{2} }$

$= \sqrt{2}$

DE

$= \sqrt{ {(3 - 5)}^{2} + {(1 + 1}^{2} } \\ = \sqrt{ {( - 2)}^{2} + {(2)}^{2} } \\ = \sqrt{4 + 4} \\ = \sqrt{8} \\ = 2 \sqrt{2}$

So DE = 2 × AB

and since the new triangle formed is similar to the original one, their side ratio will be same for all sides.

scale factor = AB/DE

= 2

It's been reflected across the Y-axis

moved thru the translation of 3 units towards the right of positive x- axis 

for this let's compare the location of points B and D

For both the y coordinate is same while the x coordinate of B is 0 and that of D is 3

so the triangle has been shifted by 3 units across the positive x axis

7 0

2 years ago

The circle below is centered at the origin, and the length of its radius is 7. What is the circle's equation. A. 7x+7y=49. B. x^

Scilla [17]

Equation of a circle is: (x−h)2+(y−k)2=r2 h is the x-coordinate of the center and k is the y-coordinate of the center. Even though the formula has minus signs, those are positive numbers. If there were plus signs in there, the numbers h and k would actually be negative. Now since you're centered at the origin, that means h and k are 0, so you'd just have this: x2+y2=r2 Now since your radius is 7, just square it and replace r^2 woth that value :)

6 0

3 years ago

Read 2 more answers

Whats the answer giving brainliest :))))) lol

n200080 [17]

Answer:

B

Step-by-step explanation:

1+1=2

But 2x has to stay the same

5 0

2 years ago

Prove that if {x1x2.......xk}isany

Radda [10]

Answer:

See the proof below.

Step-by-step explanation:

What we need to proof is this: "Assuming X a vector space over a scalar field C. Let X= {x1,x2,....,xn} a set of vectors in X, where $n\geq 2$ . If the set X is linearly dependent if and only if at least one of the vectors in X can be written as a linear combination of the other vectors"

Proof

Since we have a if and only if w need to proof the statement on the two possible ways.

If X is linearly dependent, then a vector is a linear combination

We suppose the set $X= (x_1, x_2,....,x_n)$ is linearly dependent, so then by definition we have scalars $c_1,c_2,....,c_n$ in C such that:

$c_1 x_1 +c_2 x_2 +.....+c_n x_n =0$

And not all the scalars $c_1,c_2,....,c_n$ are equal to 0.

Since at least one constant is non zero we can assume for example that $c_1 \neq 0$ , and we have this:

$c_1 v_1 = -c_2 v_2 -c_3 v_3 -.... -c_n v_n$

We can divide by c1 since we assume that $c_1 \neq 0$ and we have this:

$v_1= -\frac{c_2}{c_1} v_2 -\frac{c_3}{c_1} v_3 - .....- \frac{c_n}{c_1} v_n$

And as we can see the vector $v_1$ can be written a a linear combination of the remaining vectors $v_2,v_3,...,v_n$ . We select v1 but we can select any vector and we get the same result.

If a vector is a linear combination, then X is linearly dependent

We assume on this case that X is a linear combination of the remaining vectors, as on the last part we can assume that we select $v_1$ and we have this:

$v_1 = c_2 v_2 + c_3 v_3 +...+c_n v_n$