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2 years ago

5

Needing help with this very confusing!

2 answers:

Wittaler [7]2 years ago

6 0

Answer:

The answer is part C

Step-by-step explanation:

AveGali [126]2 years ago

3 0

Answer:

its c

Step-by-step explanation:

Ive had this question before

please give me brainliest if this helps

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If AB is 12, what is the length of A'B'?

lukranit [14]

Triangles ABC and A'B'C are similar. This means that the corresponding sides are in the same ratio, in this case:

$\frac{BC}{B^{\prime}C}=\frac{AC}{A^{\prime}C}=\frac{AB}{A^{\prime}B^{\prime}}$

Replacing with data:

$\begin{gathered} \frac{9}{6}=\frac{12}{A^{\prime}B^{\prime}} \\ 9\cdot A^{\prime}B^{\prime}=12\cdot6 \\ A^{\prime}B^{\prime}=\frac{72}{9} \\ A^{\prime}B^{\prime}=8 \end{gathered}$

7 0

6 months ago

Please help me with my hw

stellarik [79]

Part 2 is 12.5 cm represents 5km
I worked this out by doing a ratio
5cm=2km
?cm=5km
you have to do 2 times 2.5 to get 5km so you do 5 times 2.5 to get 12.5cm
so the answer is that 12.5cm

6 0

2 years ago

A _____________________ variable _________________ on an ____________________ variable.

Mkey [24]

A dependent variable depends on an independent variable

7 0

2 years ago

Help plzzzzzzzz this is hard

Yakvenalex [24]

(A) Vertical angles are congruent

6 0

2 years ago

let x1,x2, and x3 be linearly independent vectors in R^(n) and let y1=x2+x1; y2=x3+x2; y3=x3+x1. are y1,y2,and y3 linearly indep

Nutka1998 [239]

Answer with Step-by-step explanation:

We are given that

$x_1,x_2$ and $x_3$ are linearly independent.

By definition of linear independent there exits three scalar $a_1,a_2$ and $a_3$ such that

$a_1x_1+a_2x_2+a_3x_3=0$

Where $a_1=a_2=a_3=0$

$y_1=x_2+x_1,y_2=x_3+x_2,y_3=x_3+x_1$

We have to prove that $y_1,y_2$ and $y_3$ are linearly independent.

Let $b_1,b_2$ and $b_3$ such that

$b_1y_1+b_2y_2+b_3y_3=0$

$b_1(x_2+x_1)+b_2(x_3+x_2)+b_3(x_3+x_1)=0$

$b_1x_2+b_1x_1+b_2x_3+b_2x_2+b_3x_3+b_3x_1=0$

$(b_1+b_3)x_1+(b_2+b_1)x_2+(b_2+b_3)x_3=0$

$b_1+b_3=0$

$b_1=-b_3$ ...(1)

$b_1+b_2=0$

$b_1=-b_2$ ..(2)

$b_2+b_3=0$

$b_2=-b_3$ ..(3)

Because $x_1,x_2$ and $x_3$ are linearly independent.

From equation (1) and (3)

$b_1=b_2$ ...(4)

Adding equation (2) and (4)

$2b_1==0$

$b_1=0$

From equation (1) and (2)

$b_3=0,b_2=0,b_3=0$

Hence, $y_1,y_2$ and $y_3$ area linearly independent.

5 0

2 years ago