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Elena L [17]
2 years ago
7

Donna earns $1,404 per week at her job. However, 19% of her income gets taken out in taxes, and 12% of her income gets taken out

and put into her retirement account. Which is a reasonable estimate of the amount Donna is left with each week after the money is deducted?
Mathematics
1 answer:
balu736 [363]2 years ago
8 0

Answer:

About 969 dollars

Step-by-step explanation:

I kind of did it the long way. I divided $1,404 by 100 and multiplied by 19. Did the same thing with 12. Added the product of those together and subtracted from 1,404.

(1,404÷100)19 + (1,404÷100)12 = 435.24

1,404 - 435.24 = 968.76

Since it was an estimate I rounded it to the nearest whole number and got 969

Hope this helps. There are probably better methods to solve this problem but this was my way.

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Triangle P Q R is shown. Angle Q P R is a right angle. The length of Q P is 8 StartRoot 3 EndRoot and the length of P R is 8.
Vika [28.1K]

Answer: Length of side QR is 16 units.

Step-by-step explanation: Given that dimensions of  PQR

QPR= 90degrees

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2 years ago
What is AE ? Enter your answer in the box. units Two segments A D and B C intersect at point E to form two triangles A B E and D
spin [16.1K]

Answer:

The length of AE is 20 units.

Step-by-step explanation:

Given two segments AD and BC intersect at point E to form two triangles ABE and DCE. Side AB is parallel to side DC. A E is labeled 2x+10. ED is labeled x+3. AB is 10 units long and DC is 4 units long.

we have to find the length of AE

AB||CD ⇒ ∠EAB=∠EDC and ∠EBA=∠ECD  

In ΔABE and ΔDCE

∠EAB=∠EDC      (∵Alternate angles)

∠EBA=∠ECD      (∵Alternate angles)

By AA similarity, ΔABE ≈ ΔDCE

therefore, \frac{AE}{ED}=\frac{AB}{CD}

⇒ \frac{2x+10}{x+3}=\frac{10}{4}

⇒ 8x+40=10x+30

⇒ x=5

Hence, AE=2x+10=2(5)+10=20 units

The length of AE is 20 units.

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Extend Thinking: Find the x and y intercepts. (Use 2.1 & 2.2 notes to solve.)
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3 years ago
Determine formula of the nth term 2, 6, 12 20 30,42​
nalin [4]

Check the forward differences of the sequence.

If \{a_n\} = \{2,6,12,20,30,42,\ldots\}, then let \{b_n\} be the sequence of first-order differences of \{a_n\}. That is, for n ≥ 1,

b_n = a_{n+1} - a_n

so that \{b_n\} = \{4, 6, 8, 10, 12, \ldots\}.

Let \{c_n\} be the sequence of differences of \{b_n\},

c_n = b_{n+1} - b_n

and we see that this is a constant sequence, \{c_n\} = \{2, 2, 2, 2, \ldots\}. In other words, \{b_n\} is an arithmetic sequence with common difference between terms of 2. That is,

2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2

and we can solve for b_n in terms of b_1=4:

b_{n+1} = b_n + 2

b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2

b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2

and so on down to

b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)

We solve for a_n in the same way.

2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)

Then

a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)

a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))

a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))

and so on down to

a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2

\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}

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