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cestrela7 [59]
2 years ago
12

Help me please if i dont do my homework my mom will beat me

Mathematics
2 answers:
WINSTONCH [101]2 years ago
8 0

Answer:

i think its 3.5

Step-by-step explanation:

you divide 21 by 6 i think, and that should give you what its increasing by

Snezhnost [94]2 years ago
4 0

3.5 is the answer

I did the math and that is what i got.

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Mr. Thompson took his 5 children
Semmy [17]

Answer:

what is the question

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
Simplify the expression below.
densk [106]
We have:
 (2x ^ 3-12x ^ 2 + 10x-108) / x-6
 Dividing the expression between x-6 we have:
 quotient: 2x ^ 2 + 10
 remainder: - 48
 We can check the result then rewriting the expression:
 2x ^ 3-12x ^ 2 + 10x-108
 = (2x ^ 2 + 10) * (x-6) - 48
 = (2x ^ 3 + 10x - 12 x ^ 2 - 60) - 48
 = 2x ^ 3-12x ^ 2 + 10x-108
 Division OK.
 Answer:
 The result of the division is:
 quotient: 2x ^ 2 + 10
 remainder: - 48


4 0
3 years ago
tickets of a program at college cost $4 for general admission or $3 with a student ID. if 180 people paid to see a performance a
Licemer1 [7]

Answer:

75 general

105 student ID

Step-by-step explanation:

(it's a system)

4x+3y=615

x+y=180

(system as well)

4x+3y=615

x=180-y

4(180-y)+3y=615

720-4y+3y=615

-y=-105

y=105

x+105=180

x=180-105

x=75

7 0
2 years ago
Find the general term of {a_n}
Assoli18 [71]

From the given recurrence, it follows that

a_{n+1} = 2a_n + 1 \\\\ a_{n+1} = 2(2a_{n-1} + 1) = 2^2a_{n-1} + 1 + 2 \\\\ a_{n+1} = 2^2(2a_{n-2}+1) + 1 + 2 = 2^3a_{n-2} + 1 + 2 + 2^2 \\\\ a_{n+1} = 2^3(2a_{n-3} + 1) + 1 + 2 + 2^2 = 2^4a_{n-3} + 1 + 2 + 2^2 + 2^3

and so on down to the first term,

a_{n+1} = 2^na_1 + \displaystyle \sum_{k=0}^{n-1}2^k

(Notice how the exponent on the 2 and the subscript of <em>a</em> in the first term add up to <em>n</em> + 1.)

Denote the remaining sum by <em>S</em> ; then

S = 1 + 2 + 2^2 + \cdots + 2^{n-1}

Multiply both sides by 2 :

2S = 2 + 2^2 + 2^3 + \cdots + 2^n

Subtract 2<em>S</em> from <em>S</em> to get

S - 2S = 1 - 2^n \implies S = 2^n - 1

So, we end up with

a_{n+1} = 4\cdot2^n + S \\\\ a_{n+1} = 2^2\cdot2^n + 2^n-1 \\\\ a_{n+1} = 2^{n+2} + 2^n - 1 \\\\\implies \boxed{a_n = 2^{n+1} + 2^{n-1} - 1}

5 0
2 years ago
Solve each quadratic equation. Show your work. 14. (2x – 1)(x + 7) = 0 15. x2 + 3x = 10 16. 4x2 = 100
andreyandreev [35.5K]
14.  (2x - 1)(x + 7) = 0

Using the zero factor property, we know that either the first or second terms (or both) must be equal to 0 if their product is 0.  We can set each term equal to 0 to find the solutions:

2x - 1 = 0
2x = 1
x = 1/2

x + 7 = 0
x = -7

15.  x^{2} +3x=10

To solve this equation, you first need to set it equal to 0:

x^{2} +3x=10 \\ x^{2} +3x-10 = 0

Next, it can be factored:

x^{2} +3x-10 = 0 \\ (x+5)(x-2)=0

Finally, we can solve just like we did above:

x + 5 = 0
x = -5

x - 2 = 0
x = 2

16.  4x^2=100

First, you can simplify by dividing each side by 4:

4x^2=100 \\ x^2=25

Now, set the equation equal to 0:

x^2=25 \\ x^2-25=0

Next, factor:

x^2-25=0 \\ (x+5)(x-5)=0

Finally, find the solutions:

x + 5 = 0
x = -5

x - 5 = 0
x = 5
6 0
3 years ago
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