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Elena-2011 [213]

3 years ago

6

Which of the following can be used to prove two lines crossed by a transversal are parallel? Question 11 options: Congruent Alte

rnate Interior Angles Congruent Corresponding Angles Supplementary Same-Side Exterior Angles All of the above None of the above

1 answer:

beks73 [17]3 years ago

3 0

Supplementary angles

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Figure this out for 15 points

garri49 [273]

Step-by-step explanation:

Joe's climbing rate is not constant

3 × 2 = 6

8 × 1.5 = 12

12 × 1.1666666667 = 14

Time when climber reaches 30ft is unknown because climbing rate is not constant

Zas's climbing rate is 2.5

2 × 2.5 = 5

4 × 2.5 = 10

12 × 2.5 = 30

At 12 seconds, Zas reaches 30 feet.

Stephen's climbing rate is 4

2 × 4 = 8

4 × 4 = 16

6 × 4 = 24

30 ÷ 4 = 7.5

7.5 × 4 = 30

At 7.5 seconds, Stephen reaches 30 feet.

3 0

3 years ago

Matt and Alex are walking home from school one day and Alex says he knows a short cut. Matt says that the normal way they go is

nataly862011 [7]

Answer:mat

Step-by-step explanation:300 is less then 400

3 0

3 years ago

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To find the difference of 7 - 3 5/12 how do you rename the 7

lawyer [7]

You would want to make it so it has 12 on the bottom. So you would multiply by 12. Which is 84. The you put it over 12. So it would be 84/12.

7 0

3 years ago

Divide 75 by -25.<br> 12

enot [183]

Answer: 6.25

Step-by-step explanation:

6 0

3 years ago

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Let $f(x)$ be a function defined for all positive real numbers satisfying the conditions $f(x) > 0$ for all $x > 0$ and $f

algol [13]

Suppose we choose $x=1$ and $y=\dfrac12$ . Then

$f(x-y)=\sqrt{f(xy)+1}\implies f\left(\dfrac12\right)=\sqrt{f\left(\dfrac12\right)+1}\implies f\left(\dfrac12\right)=\dfrac{1+\sqrt5}2$

Now suppose we choose $x,y$ such that

$\begin{cases}x-y=\dfrac12\\\\xy=2009\end{cases}$

where we pick the solution for this system such that $x>y>0$ . Then we find

$\dfrac{1+\sqrt5}2=\sqrt{f(2009)+1}\implies f(2009)=\dfrac{1+\sqrt5}2$

Note that you can always find a solution to the system above that satisfies $x>y>0$ as long as $x>\dfrac12$ . What this means is that you can always find the value of $f(x)$ as a (constant) function of $f\left(\dfrac12\right)$ .

3 0

3 years ago

Read 2 more answers