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lions [1.4K]
3 years ago
8

Help with this question

Mathematics
1 answer:
ElenaW [278]3 years ago
5 0
Monthly- $15,600
biweekly- $26,000
weekly- $28,600
semimonthly- $19,200

Im pretty sure these are correct. Monthly is obviously once a month, biweekly is once every two months. Weekly is once a month and semimonthly is twice a month.
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The list shows 8 of Takisha's math quiz scores.
bonufazy [111]

Answer: 85

Step-by-step explanation:

Add all the numbers together than

divide by 8

5 0
2 years ago
6(4x+2)=3(8x+4) ??<br> please help asap
nasty-shy [4]
6(4x + 2) = 3(8x + 4)

Reorder the terms:
6(2 + 4x) = 3(8x + 4)
(2 * 6 + 4x * 6) = 3(8x + 4)
(12 + 24x) = 3(8x + 4)

Reorder the terms:
12 + 24x = 3(4 + 8x)
12 + 24x = (4 * 3 + 8x * 3)
12 + 24x = (12 + 24x)

Add '-12' to each side of the equation.
12 + -12 + 24x = 12 + -12 + 24x

Combine like terms: 12 + -12 = 0
0 + 24x = 12 + -12 + 24x
24x = 12 + -12 + 24x

Combine like terms: 12 + -12 = 0
24x = 0 + 24x
24x = 24x

Add '-24x' to each side of the equation.
24x + -24x = 24x + -24x

Combine like terms: 24x + -24x = 0
0 = 24x + -24x

Combine like terms: 24x + -24x = 0
0 = 0

Solving
0 = 0
4 0
2 years ago
Read 2 more answers
1 &gt; -1/2x + 5 solve for x
RUDIKE [14]

Answer:

x is greater than 8

Step-by-step explanation:

6 0
2 years ago
Help please! I need help with question 3. I don’t know how to start it. I’ll give brainly☺️
ikadub [295]
First you divide $75 and 0.04. And that will get you the answer.
I already did this question last year.
6 0
3 years ago
Read 2 more answers
For each of the following sequences, • Give a formula for the nth term in the sequence, • Give a recursive definition for the se
Andrew [12]

Answer:

(a)n^{th} = n

f(1) = 1

f(n) = f(n-1) + 1

(b)n^{th} = 2^{n-1}

f(1) = 1

f(n) = f(n-1) * 2

(c)n^{th} = n!

f(1) = 1

f(n) = f(n-1) * n

Step-by-step explanation:

(a) This is a sequence of consecutive number

n^{th} = n

f(1) = 1

f(n) = f(n-1) + 1

(b) This is a sequence of 2 to the power of n - 1. The next number is twice time of this number

n^{th} = 2^{n-1}

f(1) = 1

f(n) = f(n-1) * 2

(c) This is factorial sequence. Where the next number is this number multiplied by n^{th}

n^{th} = n!

f(1) = 1

f(n) = f(n-1) * n

5 0
2 years ago
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