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Anettt [7]
4 years ago
9

Let PQRS be a square piece of paper. P is folded onto R and then Q is folded onto S. The area of the resulting figure is 9 squar

e inches. Find the perimeter of square PQRS.

Mathematics
1 answer:
miss Akunina [59]4 years ago
4 0

Answer:

24 inches

Step-by-step explanation:

After folding the piece of paper, the result will be 1/4 the size of the original square. If you multiply the are of the triangle by 4, you will find that the area of the original square is 36 square inches.  The square root of 36 is 6, which shows us that the length of each side of the square is 6. To get the perimeter, multiply 6 by 4 to get 24 inches.

You might be interested in
Explain how to distribute the 13 to the terms inside the parentheses, then do the distribution. Use multiplication to simplify t
Nonamiya [84]

Answer:

52 + 13x

Step-by-step explanation:

To distribute something means to divide / share a part of something. The distributive property works similarly. The distributive property states that multiplying the total sum of two variables by a number is the same as the total sum of each product of individual variables and the number.

For example: a(b+c) = ab + ac

Therefore, in order to distribute the 13 into the terms inside the parentheses, you follow the distributive property and multiply each individual variable in the parentheses by 13 and add all the resulting products.

13(4 + x) = (13*4) + (13*x) = 52 + 13x

3 0
3 years ago
a) What is an alternating series? An alternating series is a whose terms are__________ . (b) Under what conditions does an alter
andriy [413]

Answer:

a) An alternating series is a whose terms are alternately positive and negative

b) An alternating series \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (-1)^{n-1} b_n where bn = |an|, converges if 0< b_{n+1} \leq b_n for all n, and \lim_{n \to \infty} b_n = 0

c) The error involved in using the partial sum sn as an approximation to the total sum s is the remainder Rn = s − sn and the size of the error is bn + 1

Step-by-step explanation:

<em>Part a</em>

An Alternating series is an infinite series given on these three possible general forms given by:

\sum_{n=0}^{\infty} (-1)^{n} b_n

\sum_{n=0}^{\infty} (-1)^{n+1} b_n

\sum_{n=0}^{\infty} (-1)^{n-1} b_n

For all a_n >0, \forall n

The initial counter can be n=0 or n =1. Based on the pattern of the series the signs of the general terms alternately positive and negative.

<em>Part b</em>

An alternating series \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (-1)^{n-1} b_n where bn = |an|  converges if 0< b_{n+1} \leq b_n for all n and \lim_{n \to \infty} b_n =0

Is necessary that limit when n tends to infinity for the nth term of bn converges to 0, because this is one of two conditions in order to an alternate series converges, the two conditions are given by the following theorem:

<em>Theorem (Alternating series test)</em>

If a sequence of positive terms {bn} is monotonically decreasing and

<em>\lim_{n \to \infty} b_n = 0<em>, then the alternating series \sum (-1)^{n-1} b_n converges if:</em></em>

<em>i) 0 \leq b_{n+1} \leq b_n \forall n</em>

<em>ii) \lim_{n \to \infty} b_n = 0</em>

then <em>\sum_{n=1}^{\infty}(-1)^{n-1} b_n  converges</em>

<em>Proof</em>

For this proof we just need to consider the sum for a subsequence of even partial sums. We will see that the subsequence is monotonically increasing. And by the monotonic sequence theorem the limit for this subsquence when we approach to infinity is a defined term, let's say, s. So then the we have a bound and then

|s_n -s| < \epsilon for all n, and that implies that the series converges to a value, s.

And this complete the proof.

<em>Part c</em>

An important term is the partial sum of a series and that is defined as the sum of the first n terms in the series

By definition the Remainder of a Series is The difference between the nth partial sum and the sum of a series, on this form:

Rn = s - sn

Where s_n represent the partial sum for the series and s the total for the sum.

Is important to notice that the size of the error is at most b_{n+1} by the following theorem:

<em>Theorem (Alternating series sum estimation)</em>

<em>If  \sum (-1)^{n-1} b_n  is the sum of an alternating series that satisfies</em>

<em>i) 0 \leq b_{n+1} \leq b_n \forall n</em>

<em>ii) \lim_{n \to \infty} b_n = 0</em>

Then then \mid s - s_n \mid \leq b_{n+1}

<em>Proof</em>

In the proof of the alternating series test, and we analyze the subsequence, s we will notice that are monotonically decreasing. So then based on this the sequence of partial sums sn oscillates around s so that the sum s always lies between any  two consecutive partial sums sn and sn+1.

\mid{s -s_n} \mid \leq \mid{s_{n+1} -s_n}\mid = b_{n+1}

And this complete the proof.

5 0
4 years ago
What are the solutions to the equation x^2+6=40?
Andreas93 [3]

Answer:

x = 4  or x = -10   for  equation x^2 + 6x = 40

Step-by-step explanation:

x^2 + 6 = 40

x^2 = 40 - 6

x^2 = 34

x = root(34) or  x = -root(34)   but this is not in the answer choice.

is it  x^2 + 6x = 40

x^2 + 6x + 9 = 40 +9

(x + 3)^2 = 49

x + 3 = 7  or  x + 3 = -7

x = 4  or x = -10

5 0
3 years ago
I give up i might have to pull an allnighter
netineya [11]

Answer:

Yes, they are congruent

Step-by-step explanation:

Yes, they are congruent because all the side and angle lengths are congruent.

6 0
4 years ago
Read 2 more answers
5. Identify the type and subtype of each of the fol-
r-ruslan [8.4K]

Answer:

a) Kylie has 8 marbles

b) 7 Cylinders

c) 17 carrots

d) 8 marbles belong to Shauntay

Step-by-step explanation:

5. Identify the type and subtype of each of the fol-

lowing problems.

a. Shawn has 15 marbles, which is 7 more marbles than Kyle has. How many marbles does Kyle have?

Shawn = 15 marbles

S = K + 7

15 = K + 7

K = 15 - 7

K = 8 marbles

Kylie has 8 marbles

b. Tiffany has 12 blocks, 5 of which are cubes and the rest cylinders. How many blocks are cylinders?

T = 12 blocks

Cubes = 5

Cylinders = the rest

12 blocks = Cubes + Cylinders

Cylinders = 12 - Cubes

Cylinders = 12 - 5

Cylinder = 7

c. Peter had some carrots. After he ate 3 of them, he had 14 carrots left. How many carrots did Peter have before?

Number of carrots Peter has before

= Number of carrots he ate + Number of carrots he has now

= 14 + 3

= 17 carrots

d. In a bag of 17 marbles, 9 marbles belong to Kelly and the rest belong to Shauntay. How many marbles belong to Shauntay?

Total number of Marbles = 17

Kelly = 9 marbles

Shauntay = ?

Total = Kelly + Shauntay

Shauntay = Total - Kelly's marbles

= ( 17 - 9) marbles

= 8 marbles

8 marbles belong to Shauntay

6 0
3 years ago
Read 2 more answers
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