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3 years ago

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PLZ HELP ASAP What can you say about the relations between ∠8and∠4? Explain using a basic rigid motion. Name another pair of ang

les with this same relationship.

2 answers:

Norma-Jean [14]3 years ago

6 0

Answer:

Step-by-step explanation:

In math, when it comes to angles such as 8 and 4, 7 and 3, 5 and 1, and 6 and 2, are always proportinal. This means that they are equal to eachother, because they are correspnding angles.

Tanzania [10]3 years ago

5 0

Answer:

Step-by-step explanation:

In math, when it comes to angles such as 8 and 4, 7 and 3, 5 and 1, and 6 and 2, are always proportinal. This means that they are equal to each other, because they are correspnding angles.

You might be interested in

the plans for your 13th birthday party are well under way. your parents agree to your guest list of 400 people. you decided it w

diamong [38]

Answer:

11 people will get both the prizes.

Step-by-step explanation:

In this question we need to find common integral multiples of 5 and 7 under 400. As you can see in the figure 35th person is the first one to get both ipod and psp. The LCM (least common multiple) of 5 and 7 which is 35 (!!! <em>I hope you know how to LCM of two integers </em>).

Therefore the integral multiples of 35 will be same as the common multiples of 5 and 7. So, figuring out the number of integral multiples of 35 under 400 will give the answer.

To do that divide the number 400 with 35. On performing the division the quotient will be 11 and the remainder will be 15.

Therefore 11 persons out of 400 will get both the prizes.

Similarly if you want to find the number of person getting ipod then divide 400 with 5. The quotient will be 80. ∴ 80 people will get ipod.

And for psp, divide 400 with 7. The quotient will be 57. ∴ 57 people will get psp.

In these divisions, consider just the quotient, here remainder is of no use.

5 0

3 years ago

There are 40 people at Drake's sweet sixteen party. Drake's only rule is that everyone in the party must meet and shake each oth

nikdorinn [45]

1) This is a combination, so use ⁴⁰C₂.
⁴⁰C₂ = 780
There are 780 handshakes in total.

2) I got my solution by using the combinations formula. Or another way you could do this is to add up... 39 + 38 + 37 + 36 + 35 + 34 + 33 + 32 + 31... +1. You will end up with 780 (I did this to double check my work).

3) The quickest way to figure this out if Drake invited 1,000 people is to use the natural numbers formula = n(n+ 1)/2.
Plug the numbers in: 999(1000)/2 = 499,500
If Drake invited 1,000 people, there would be 499,500 handshakes.

7 0

3 years ago

Read 2 more answers

Jeremy didn't pay his previous month's credit card bill in full. This month's bill shows a finance charge of $85.50. Jeremy calc

Nostrana [21]

Answer:

The monthly percentage rate would be 1.6%

Step-by-step explanation:

Given,

Finance charge = $ 85.50,

Amount of bill in this month = $ 5,343.75,

Hence, the monthly rate of percentage

$=\frac{\text{Finance charge}}{\text{Amount of bill}}\times 100$

$=\frac{85.5}{5343.75}\times 100$

$=\frac{8550}{5343.75}$

$=\frac{855000}{534375}$

= 1.6 %

i.e. OPTION A would be correct.

6 0

3 years ago

Use the elimination method to solve 4x+8y=16 4x-8y=0

Rina8888 [55]

Answer:

x=2, y=1

Step-by-step explanation:

4x+8y=16

4x-8y=0

Add the two equations together:

8x=16

Divide both sides by 8:

x=2

Plug this back into one of the original equations to find y:

4(2)-8y=0

8-8y=0

y=1

Hope this helps!

3 0

3 years ago

Read 2 more answers

Determine whether the sequences converge.

Alik [6]

$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

and so on. We're getting an inkling that the explicit closed form for the sequence may be $a_n=2^{(2^n-1)/2^n}$ , but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

4 0

3 years ago