X +3+ X -8+ X equals 55

PART A: Mrs. konsdorf claims that angle R is a right angle.Is Mrs. konsdorf correct? explain your reasoning PART B: if T is trab

Anna007 [38]

Answer:

Part A: Angle R is not a right angle.

Part B; Angle GRT' is a right angle.

Step-by-step explanation:

Part A:

From the given figure it is noticed that the vertices of the triangle are G(-6,5), R(-3,1) and T(2,6).

Slope formula

$m=\frac{y_2-y_1}{x_2-x_1}$

The product of slopes of two perpendicular lines is -1.

Slope of GR is

$\text{Slope of GR}=\frac{1-5}{-3-(-6)}=\frac{-4}{3}$

Slope of RT is

$\text{Slope of RT}=\frac{6-1}{2-(-3)}=\frac{5}{5}=1$

Product of slopes of GR and RT is

$\frac{-4}{3}\times 1=\frac{-4}{3}\neq -1$

Therefore lines GR and RT are not perpendicular to each other and angle R is not a right angle.

Part B:

If vertex T translated by rule

$(x,y)\rightarrow(x-1,y-2)$

Then the coordinates of T' are

$(2,6)\rightarrow(2-1,6-2)$

$(2,6)\rightarrow(1,4)$

Slope of RT' is

$\text{Slope of RT'}=\frac{4-1}{1-(-3)}=\frac{3}{4}$

Product of slopes of GR and RT' is

$\frac{-4}{3}\times \frac{3}{4}=-1$

Since the product of slopes is -1, therefore the lines GR and RT' are perpendicular to each other and angle GRT' is a right angle.

6 0

3 years ago

Estimate the quotient what is 467.6 divided by 8

lorasvet [3.4K]

I’m thinking 58.45? if that makes sense

7 0

3 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:

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