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siniylev [52]
3 years ago
13

PLS HELP IM TIMED!!

Mathematics
1 answer:
Ad libitum [116K]3 years ago
7 0

Answer:

Van 8 and bus 27

Step-by-step explanation:

v+7b=197

12v+b=123      b=123-12v

v+7(123-12v)=197

v+861-84v=197

664=83v

v=8

8+7b=197

7b=181

b=27

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Type the expression that results from the following series of steps: start with x, divide 6, then subtract 1
Alexus [3.1K]

Answer:

x/6 - 1

Step-by-step explanation:

x divide 6 subtract 1

4 0
3 years ago
Jenny and Natalie are selling cheesecakes for a school fundraiser. Customers can buy chocolate cakes and vanilla cakes. Jenny so
IrinaVladis [17]

The cost of 1 chocolate cake is $ 6 and cost of 1 vanilla cake is $ 7

<em><u>Solution:</u></em>

Let "c" be the cost of 1 chocolate cake

Let "v" be the cost of 1 vanilla cake

<em><u>Jenny sold 14 chocolate cakes and 5 vanilla cakes for 119 dollars</u></em>

Therefore, we can frame a equation as:

14 x cost of 1 chocolate cake + 5 x cost of 1 vanilla cake = 119

14 \times c + 5 \times v=119

14c + 5v = 119 ------- eqn 1

<em><u>Natalie sold 10 chocolate cakes and 10 vanilla cakes for 130 dollars</u></em>

Therefore, we can frame a equation as:

10 x cost of 1 chocolate cake + 10 x cost of 1 vanilla cake = 130

10 \times c + 10 \times v = 130

10c + 10v = 130 -------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

Multiply eqn 1 by 2

28c + 10v = 238 ------ eqn 3

<em><u>Subtract eqn 2 from eqn 3</u></em>

28c + 10v = 238

10c + 10v = 130

( - ) --------------------------

18c = 108

c = 6

<em><u>Substitute c = 6 in eqn 1</u></em>

14(6) + 5v = 119

84 + 5v = 119

5v = 119 - 84

5v = 35

v = 7

Thus cost of 1 chocolate cake is $ 6 and cost of 1 vanilla cake is $ 7

8 0
3 years ago
Please answer!!!!!!
Crazy boy [7]

Answer:

(this isnt a mathematics question, it's science, btw)

Indeed, you are correct. It is 1. I think.

Because you get one allele from each parent, to make up 2 alleles per offspring.

Hope that helps!

4 0
3 years ago
The area of a regular octagon is 35 cm 2 what is the area of a regular octagon with sides three times as large
Maurinko [17]
The area of a polygon = n*side^2 / (4 * tan(180/n))
where "n" is the number of sides

Let's calculate area for side = 2
area = 8*2^2 / (4 * tan(180/8))
area = 32 / (4 * tan(22.5))
area = 32 / 4*0.41421
<span> <span> <span> area = 19.3138746047 </span> </span> </span>

Now, let's calculate area for side = 6
area = 8*6^2 / (4 * tan(180/8))
area = 288 / <span> <span> <span> 1.65684 </span> </span> </span>
area = <span> <span> <span> 173.824871442 </span> </span> </span>

<span> <span> 173.824871442 </span> / </span><span><span>19.3138746047 = </span> 9

So, the area would be 9 times larger.

ALSO, looking at the formula
</span>n*side^2 / (4 * tan(180/n))
we can see that the side length appears just once in the formula and we are to square it in the calculation.  So, if we increase the side length is increased by 3, then the area increases by 3^2 or 9.


5 0
4 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}

6 0
2 years ago
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