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never [62]
3 years ago
10

This is geometry. Please help explain this.

Mathematics
1 answer:
lina2011 [118]3 years ago
5 0
Sorry I don’t now Gemini
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Nikki has $1100 and spends $35 per week. Mich has $520 and saves $55 per week. In about how many weeks, will Nikki and Micah hav
Rashid [163]

Answer:

6 weeks

Step-by-step explanation:

Given that :

Nikki :

Initial amount = $1100 ;

Amount spent per week = $35

Mich:

Initial amount = $520

Amount saved per week = $55

Number of weeks they'll both have the same amount :

Let number of weeks = w

For Nikki who is spending :

1100 - 35w - - - (1)

For Mich who is saving :

520 + 55w - - - (2)

Equate (1) and (2)

1100 - 35w = 520 + 55w

1100 - 520 = 55w + 35w

580 = 90w

w = 580 / 90

w = 6.444

About 6 weeks

5 0
2 years ago
Fran baked 300 cookies for her bakery.15% of the cookies were oatmeal raisin.25% were snickerdoodle and the rest were chocolate
Masteriza [31]

Answer:

180

Step-by-step explanation:

15% of 300 is 45.

25% of 300 is 75.

45+75 is 120.

300-120 is 180

5 0
3 years ago
Select the true statement plzzzz need help ASAPPPPP
xxMikexx [17]

Answer:

Correct choice is B

Step-by-step explanation:

All three functions are increasing when x is increasing. You can find values of each function when x is "enough large". For example, at x=100,

1. linear function: y=10\cdot 100=1,000.

2. exponential function: y=5^{100}\approx 0.8\cdot 10^{70}.

3. quadratic function: y=4\cdot 100^2+5\cdot 100=40,500.

As you can see,  when x approaches positive infinity, the exponential function will exceed both the linear and the quadratic functions.

Also you can use graphical method to get from the attached diagrams result. When x approaches positive infinity, the exponential function will exceed both the linear and the quadratic functions.

Correct choice is B.

5 0
3 years ago
How do i find the value of x?
inn [45]

Answer:h

2x+1+3x+8./2=. 17

5x+9. = 17 x2

5x+9= 34

-9. -9

5x=25

Divide by 5

X=5

Step-by-step explanation:

4 0
3 years ago
What is a quick and easy way to remember explicit and recursive formulas?
Oliga [24]
I always found derivation to be helpful in remembering. Since this question is tagged as at the middle school level, I assume you've only learned about arithmetic and geometric sequences.

First, remember what these names mean. An arithmetic sequence is a sequence in which consecutive terms are increased by a fixed amount; in other words, it is an additive sequence. If a_n is the nth term in the sequence, then the next term a_{n+1} is a fixed constant (the common difference d) added to the previous term. As a recursive formula, that's

a_{n+1}=a_n+d

This is the part that's probably easier for you to remember. The explicit formula is easily derived from this definition. Since a_{n+1}=a_n+d, this means that a_n=a_{n-1}+d, so you plug this into the recursive formula and end up with 

a_{n+1}=(a_{n-1}+d)+d=a_{n-1}+2d

You can continue in this pattern, since every term in the sequence follows this rule:

a_{n+1}=a_{n-1}+2d
a_{n+1}=(a_{n-2}+d)+2d
a_{n+1}=a_{n-2}+3d
a_{n+1}=(a_{n-3}+d)+3d
a_{n+1}=a_{n-3}+4d

and so on. You start to notice a pattern: the subscript of the earlier term in the sequence (on the right side) and the coefficient of the common difference always add up to n+1. You have, for example, (n-2)+3=n+1 in the third equation above.

Continuing this pattern, you can write the formula in terms of a known number in the sequence, typically the first one a_1. In order for the pattern mentioned above to hold, you would end up with

a_{n+1}=a_1+nd

or, shifting the index by one so that the formula gives the nth term explicitly,

a_n=a_1+(n-1)d

Now, geometric sequences behave similarly, but instead of changing additively, the terms of the sequence are scaled or changed multiplicatively. In other words, there is some fixed common ratio r between terms that scales the next term in the sequence relative to the previous one. As a recursive formula,

a_{n+1}=ra_n

Well, since a_n is just the term after a_{n-1} scaled by r, you can write

a_{n+1}=r(ra_{n-1})=r^2a_{n-1}

Doing this again and again, you'll see a similar pattern emerge:

a_{n+1}=r^2a_{n-1}
a_{n+1}=r^2(ra_{n-2})
a_{n+1}=r^3a_{n-2}
a_{n+1}=r^3(ra_{n-3})
a_{n+1}=r^4a_{n-3}

and so on. Notice that the subscript and the exponent of the common ratio both add up to n+1. For instance, in the third equation, 3+(n-2)=n+1. Extrapolating from this, you can write the explicit rule in terms of the first number in the sequence:

a_{n+1}=r^na_1

or, to give the formula for a_n explicitly,

a_n=r^{n-1}a_1
6 0
3 years ago
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