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kakasveta [241]

1 year ago

9

Given: x ∥ y and w is a transversal Prove: ∠3 ≅ ∠6 Parallel lines x and y are cut by transversal w. On line x where it intersect

s with line w, 4 angles are created. Labeled clockwise, from uppercase left, the angles are: 2, 4, 3, 1. On line y where it intersects with line w, 4 angles are created. Labeled clockwise, from uppercase left, the angles are: 6, 8, 7, 5. What is the missing reason in the proof? Statement Reason 1. x ∥ y w is a transversal 1. given 2. ∠2 ≅ ∠3 2. def. of vert. ∠s 3. ∠2 ≅ ∠6 3. def. of corr. ∠s 4. ∠3 ≅ ∠6 4. transitive property symmetric property vertical angles are congruent definition of supplementary angles

2 answers:

alisha [4.7K]1 year ago

7 0

Answer:

A.

Step-by-step explanation:

SOVA2 [1]1 year ago

6 0

Answer:

The missing reason in the proof is transitive property

Step-by-step explanation:

<u>Statement </u> <u>Reason </u>

1. x ∥ y w is a transversal 1. given

2. ∠2 ≅ ∠3 2. def. of vert. ∠s

3. ∠2 ≅ ∠6 3. def. of corr. ∠s

4. ∠3 ≅ ∠6 4. ??????????

From the statements 2 and 3

The previous proved statement to make use of the transitive property reason or proof

∴ 4. ∠3 ≅ ∠6 4. transitive property

Note: the transitive property states that: If a = b and b = c, then a = c.

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Show that the Fibonacci numbers satisfy the recurrence relation fn = 5fn−4 + 3fn−5 for n = 5, 6, 7, . . . , together with the in

Sonja [21]

Answer with step-by-step explanation:

We are given that the recurrence relation

$f_n=5f_{n-4}+3f_{n-5}$

for n=5,6,7,..

Initial condition

$f_0=0,f_1=1,f_2=1,f_3=2,f_4=3$

We have to show that Fibonacci numbers satisfies the recurrence relation.

The recurrence relation of Fibonacci numbers

$f_n=f_{n-1}+f_{n-2}$ , $f_0=0,f_1=1$

Apply this

$f_n=(f_{n-2}+f_{n-3})+f_{n-2}=2f_{n-2}+f_{n-3}$

$f_n=2(f_{n-3}+f_{n-4})+f_{n-3}=3f_{n-3}+2f_{n-4}$

$f_n=3(f_{n-4}+f_{n-5})+2f_{n-4}=5f_{n-4}+3f_{n-5}$

Substitute n=2

$f_2=f_1+f_0=1+0=1$

$f_3=f_2+f_1=1+1=2$

$f_4=f_3+f_2=2+1=3$

Hence, the Fibonacci numbers satisfied the given recurrence relation .

Now, we have to show that $f_{5n}$ is divisible by 5 for n=1,2,3,..

Now replace n by 5n

$f_{5n}=5f_{5n-4}+3f_{5n-5}$

Apply induction

Substitute n=1

$f_5=5f_1+3f_0=5+0=5$

It is true for n=1

Suppose it is true for n=k

$f_{5k}=5f_{5k-4}+3f_{5k-5}$ is divisible 5

Let $f_{5k}=5q$

Now, we shall prove that for n=k+1 is true

$f_{5k+5}=5f_{5k+5-4}+3f_{5k+5-5}=5f_{5k+1}+3f_{5k}=5f_{5k+1}+3(5q)$

$f_{5k+5}=5(f_{5k+1}+3q)$

It is multiple of 5 .Therefore, it is divisible by 5.

It is true for n=k+1

Hence, the $f_{5n}$ is divisible by 5 for n=1,2,3,..

8 0

2 years ago

A ranger in a lookout tower spots two fires on the campground below. Fire A is 75 meters east

Wittaler [7]

Answer:

the fire hydrant is 9 meters west and 58 m south of the tower

Step-by-step explanation:

Fire A and Fire B can be represented as a point in the Cartesian coordinate with the tower as the origin. Fire A is 75 meters east and 40 meters south of the tower, It can be represented as (75, -40). Fire B is 37 meters west and 64 meters south of the tower, It can be represented as (-37, -64).

If a point O(x, y) divides a line segment AB in the ratio n:m, with A( $x_1,y_1$ ) and B( $x_2,y_2$ ), the point O(x, y) is at:

$x=\frac{n}{n+m}(x_2-x_1)+x_1\\ \\y=\frac{n}{n+m}(y_2-y_1)+y_1$

The fire hydrant (x, y) divides Fire A(75, -40) and fire B(-37, -64) in the ratio 3:1(three fourth), hence:

$x=\frac{n}{n+m}(x_2-x_1)+x_1=\frac{3}{4}(-37-75) +75=\frac{3}{4}(-112)+75=-84+75\\x=-9 \\ \\y=\frac{n}{n+m}(y_2-y_1)+y_1=\frac{3}{4}(-64-(-40))+(-40)= \frac{3}{4}(-18)-40=-18-40\\y=-58$

The fire hydrant is at (-9, -58). That is the fire hydrant is 9 meters west and 58 m south of the tower

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1 year ago

For the equation ae^ct=d, solve for the variable t in terms of a,c, and d. Express your answer in terms of the natural logarithm

saveliy_v [14]

We have been given an equation $ae^{ct}=d$ . We are asked to solve the equation for t.

First of all, we will divide both sides of equation by a.

$\frac{ae^{ct}}{a}=\frac{d}{a}$

$e^{ct}=\frac{d}{a}$

Now we will take natural log on both sides.

$\text{ln}(e^{ct})=\text{ln}(\frac{d}{a})$

Using natural log property $\text{ln}(a^b)=b\cdot \text{ln}(a)$ , we will get:

$ct\cdot \text{ln}(e)=\text{ln}(\frac{d}{a})$

We know that $\text{ln}(e)=1$ , so we will get:

$ct\cdot 1=\text{ln}(\frac{d}{a})$

$ct=\text{ln}(\frac{d}{a})$

Now we will divide both sides by c as:

$\frac{ct}{c}=\frac{\text{ln}(\frac{d}{a})}{c}$

$t=\frac{\text{ln}(\frac{d}{a})}{c}$

Therefore, our solution would be $t=\frac{\text{ln}(\frac{d}{a})}{c}$ .

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Xavier is a salesperson who is paid a fixed amount of $455 per week. He also earns a commission of 3% on the sales he makes. If

melomori [17]

Xavier wants to earn more then 575 dollars a week, he will have to sale x<3,600 a week . So the answer is C

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nevsk [136]

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