All Answers

#1

+129839

+1

Intersecting Chord Theorem

Let CQ = x and DQ = 38 - x

We have that

AQ * BQ = CQ * DQ

6 * 12 = x ( 38 - x)

72 = 38x - x^2 rearrange as

x^2 - 38x + 72 = 0 factor as

(x - 36) ( x - 2) = 0

Setting each factor to 0 and solving for x we have

x = 36 reject

x = 2 = min length of CQ

CPhillMay 18, 2021

#1

+1

You are subtracting a negative, so make it positive, and then add the two numbers, giving you a sum of 21.06

GuestMay 18, 2021

#4

+197

+1

Let m be the total number of boys in the family, and f the total number of girls. Simon has m-1 brothers and f sisters, while Sophia has f-1 sisters and m brothers.

For Simon,

m-1 = f ==> m = f+1

For Sophia,

m = 2*(f-1) ==> m = 2*f - 2

Replacing m with f+1:

f + 1 = 2*f - 2 ==> f = 3

So m = 4, f = 3 from the first equation

Checking our work:

For Simon, 4–1 = 3; for Sophia, 4 = 2*(3–1)

TLDR: four boys and three girls in the family

KINGCASHMay 18, 2021

#3

+54

+1

1. If we set b to the number of total brothers, and s to the total number of sisters, we can write and solve the following system:

b - 1 = s (because the number of total brothers, minus Simon is equal to the number of sisters)

2(s - 1) = b (because twice the number of sisters, minus Sophia is equal to the number of brothers)

Solving for this gives us a total of 4 brothers and 3 sisters, meaning there are 7 total children (assuming all the brothers and sisters are children)

hope this helps (sorry if it took too long)

mitgliedd1May 18, 2021

#1

+313

+1

https://web2.0calc.com/questions/m-a-t-h_5

I answered it here. I hope this helps

apsiganocjMay 18, 2021

#1

+515

+2

$24-2x>17$

$ x<3.5$

so numbers 3, 2, 1

3 + 2 + 1 = 6

mworkhard222May 18, 2021

#3

+313

0

Agreed

apsiganocjMay 18, 2021

#1

+313

+1

20¢ 50¢
Box A : Box B Diff Box B -- Box A

2 : 1 0.50--0.40 = $1.60

0.20 * 2 : 0.50 0.104 = 1.60

0.40¢ u = 1.60
------

16, 0.40¢ Coins Box A 0.10

32, 0.20¢ Coins Box A u = 16

The answer choice is C

apsiganocjMay 18, 2021

#2

+129839

0

Very nice, mworkhard !!!!

CPhillMay 18, 2021

#2

+515

+2

thanks, just helped u on ur question too!

mworkhard222May 18, 2021

#1

+515

+3

let x = # of red and blue stickers

2x = total stickers

$2x-\frac23x-125=\frac14\cdot2x$

$2x-\frac{2}{3}x=\frac{1}{2}x+125$

$\frac{5}{6}x=125$

$5x=750$

$x=150$

Question #1: $\frac{125}{150}=\frac{5}{6}$

Question #2: 150

mworkhard222May 18, 2021

#4

+129839

+1

I'll do the first one..you should be able to do the rest..we will use the "divide and average" method

Let us suppose that we think 3 is the sq rt of 20

1. Divide 20 by 3 = 6.66

2. Average 3 + 6.66 = (3 + 6.66) / 2 = 4.83

3. Divide 20 by 4.83 = 4.14

4. Average 4.83 and 4.14 = ( 4.83 + 4.14) / 2 = 4.485

5. divide 20 by 4.485 = 4.459

6. Avverage 4.485 and 4.459 = 4.472

7. Divide 20 by 4.472 = 4.472

The last two results agree to two decimal places

So.....sqrt 20 ≈ 4.47

CPhillMay 18, 2021

#1

+54

+1

2. 9 * 3 = 27

27 buckets

i will be back with more

mitgliedd1May 18, 2021

#1

+515

+1

1.

a) 7

b) 11

2. √250 $\approx$ 15.81
15.81 * 4 = 63.24
So 63

mworkhard222May 18, 2021

#1

+129839

+1

Paul had 3/4 as many toys as Clay. When Clay gave half of his toys to Paul, he had 24 fewer toys than his friend. How many toys did Clay have at first?

Call the number that Clay had at first = N

Then Paul had (3/4)N

After Caly gave half of his toys to Clay....Clay has N /2 and Paul has (3/4)N + N/2

And Paul has 24 more than Clay....so....

N/2 + 24 = (3/4)N + N/2 simplify (subtract N/2 from both sides)

24 = (3/4)N

24 (4/3) = N = 32 = the number that Clay had at first

CPhillMay 18, 2021

#3

+515

+1

sure, you said i could use a calculator so i did, then i rounded it to the nearest hundredth