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2 years ago

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WXYZ is a trapezoid with WX ≅ YZ. Isosceles trapezoid W X Y Z is shown. Sides X Y and W Z are parallel. The length of X Y is 3 a

minus 4 and the length of W Z is 3 a. Sides W X and Y Z are congruent. The length of W X is 2 a minus 2 and the length of Y Z is a + 1. What is the length of XY? 3 units 4 units 5 units 6 units

2 answers:

mash [69]2 years ago

8 0

Answer: 5 units

Step-by-step explanation:

kykrilka [37]2 years ago

6 0

Answer:

5

Step-by-step explanation:

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What types of concurrent constructions are needed to find the orthocenter of a triangle? A. intersection of the lines drawn perp

finlep [7]

Answer: Hello! The answer to your question is B, the intersection of the lines drawn to bisect each vertex of the triangle. Hope this helped! Please pick my answer as the Brainliest!

4 0

2 years ago

If you take out a $15,000 loan for a period of 36 months at a 5.4% annual interest rate, how much of your first monthly payment

Tems11 [23]

Answer:

The amount that will go towards the balance is 385.71.

Step-by-step explanation:

The information provided is:

P = Principal of loan = $15,000

r = interest rate = 5.4% p.a. = $\frac{0.054}{12}= 0.0045$

n = number of periods = 36

The amount of the monthly payment that will go towards the balance is:

Amount towards the balance = EMI - Interest

First compute the Equated monthly installments (EMI) as follows:

$EMI=\frac{P\times r\times (1+r)^{n}}{(1+r)^{n}-1} \\=\frac{15000\times0.0045\times(1+0.0045)^{36}}{(1+0.0045)^{36}-1} \\=\frac{79.3125}{0.175}\\=453.21$

Now compute the interest as follows:

$Interest=P\times r\\=15000\times0.0045\\=67.5$

The amount that will go towards the balance is,

Amount towards the balance = EMI - Interest

= 453.21 - 67.5

= 385.71

Thus, the amount that will go towards the balance is 385.71.

4 0

2 years ago

Read 2 more answers

LESSON 1 SESSION 1

denpristay [2]

Answer:

<em>Between which two tens does it fall?</em><em> </em><u>Between 25 and 26 tens</u>

<em><u /></em>

<em>Between which two hundreds does it fall?</em> <u>Between 2 and 3 hundreds</u>

Explanation:

The place-value chart is:

Hundreds Tens Ones

2 5 3

<em><u /></em>

<em><u>a) Between which two tens does it fall? </u></em>

Using the place values you can write 253 = 25 × 10 + 3, i.e. 25 tens and 3 ones.

From that you can write:

250 < 253 < 260

250 = 25 × 10 = 25 tens

260 = 26 × 10 = 26 tens

Then, you conclude that 253 is between 25 and 26 tens.

<u><em>b) Between which two hundreds does it fall?</em></u>

Using the same reasoning:

253 = 2 × 100 + 5 × 10 + 3 = 253

200 < 253 < 300

200 = 2 hundreds

300 = 3 hundreds

Conclusion: 253 is between 2 hundreds and 3 hundreds.

3 0

2 years ago

Please help me and I’ll give you brainiest!!!!!

algol [13]

Answer:

D

Step-by-step explanation:

8 0

2 years ago

A store sells 8 colors of balloons with at least 28 of each color. How many different combinations of 28 balloons can be chosen?

Len [333]

Answer:

(a) $Selection = 6724520$

(b) $At\ most\ 12 = 6553976$

(c) $At\ most\ 8 = 6066720$

(d) $At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

Step-by-step explanation:

Given

$Colors = 8$

$Balloons = 28$ --- at least

Solving (a): 28 combinations

From the question, we understand that; a combination of 28 is to be selected. Because the order is not important, we make use of combination.

Also, because repetition is allowed; different balloons of the same kind can be selected over and over again.

So:

$n => 28 + 8-1$ $= 35$

$r = 28$

$Selection = ^{35}^C_{28$

$Selection = \frac{35!}{(35 - 28)!28!}$

$Selection = \frac{35!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29*28!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29}{7!}$

$Selection = \frac{35*34*33*32*31*30*29}{7*6*5*4*3*2*1}$

$Selection = \frac{33891580800}{5040}$

$Selection = 6724520$

Solving (b): At most 12 red balloons

First, we calculate the ways of selecting at least 13 balloons

Out of the 28 balloons, there are 15 balloons remaining (i.e. 28 - 13)

So:

$n => 15 + 8 -1 = 22$

$r = 15$

Selection of at least 13 =

$At\ least\ 13 = ^{22}C_{15}$

$At\ least\ 13 = \frac{22!}{(22-15)!15!}$

$At\ least\ 13 = \frac{22!}{7!15!}$

$At\ least\ 13 = 170544$

Ways of selecting at most 12 =

$At\ most\ 12 = Total - At\ least\ 13$ --- Complement rule

$At\ most\ 12 = 6724520- 170544$

$At\ most\ 12 = 6553976$

Solving (c): At most 8 blue balloons

First, we calculate the ways of selecting at least 9 balloons

Out of the 28 balloons, there are 19 balloons remaining (i.e. 28 - 9)

So:

$n => 19+ 8 -1 = 26$

$r = 19$

Selection of at least 9 =

$At\ least\ 9 = ^{26}C_{19}$

$At\ least\ 9 = \frac{26!}{(26-19)!19!}$

$At\ least\ 9 = \frac{26!}{7!19!}$

$At\ least\ 9 = 657800$

Ways of selecting at most 8 =

$At\ most\ 8 = Total - At\ least\ 9$ --- Complement rule

$At\ most\ 8 = 6724520- 657800$

$At\ most\ 8 = 6066720$

Solving (d): 12 red and 8 blue balloons

First, we calculate the ways for selecting 13 red balloons and 9 blue balloons

Out of the 28 balloons, there are 6 balloons remaining (i.e. 28 - 13 - 9)

So:

$n =6+6-1 = 11$

$r = 6$

Selection =

$^{11}C_6 = \frac{11!}{(11-6)!6!}$

$^{11}C_6 = \frac{11!}{5!6!}$

$^{11}C_6 = 462$

Using inclusion/exclusion rule of two sets:

$Selection = At\ most\ 12 + At\ most\ 8 - (12\ red\ and\ 8\ blue)$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 170544+ 657800- 462$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = Total - Only\ 12\ red\ and\ only\ 8\ blue$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 6724520 - 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

3 0

2 years ago