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2 years ago

Please help me dunduendjenbe!!! I beg u

Mathematics

2 answers:

Aliun [14]2 years ago

6 0

Answer:

the answer is B

Step-by-step explanation:

Y=(slope)x+y-intercept/y=mx+b

plug in the slope then the Yy intercept

Harman [31]2 years ago

3 0

Answer:

y = 4/3x - 2, B is your answer

Step-by-step explanation:

Y intercept is the number next to slope(b).

The co efficient of x will be your slope so therefore the co efficient of x will be 4/3

y is always there because it will always be the dependent variable on the left side.

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Robbie has 2 times as many apples as Bob. After receiving 35 more, Robbie now has 7 times as many as Bob. How many apples does B

Hoochie [10]

Answer: 18.5 apples

Step-by-step explanation:

7 0

2 years ago

Hi can anyone solve and give a explaination plz? its for a friend

lbvjy [14]

6.75

-2.25

----------

4.50

-2.25

----------

2.25

-2.25

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0.00

you could bowl 3 times (it starts off at 6.75 because I took away the cost of the shoes)

5 0

3 years ago

Read 2 more answers

Solve the equation. 12x+3=51

Zinaida [17]

$12x + 3 = 51 \\ 12x = 48 \\ x = 4$

3 0

3 years ago

How many terms of the arithmetic sequence {1,22,43,64,85,…} will give a sum of 2332? Show all steps including the formulas used

MA_775_DIABLO [31]

There's a slight problem with your question, but we'll get to that...

Consecutive terms of the sequence are separated by a fixed difference of 21 (22 = 1 + 21, 43 = 22 + 21, 64 = 43 + 21, and so on), so the n-th term of the sequence, a (n), is given recursively by

• a (1) = 1

• a (n) = a (n - 1) + 21 … … … for n > 1

We can find the explicit rule for the sequence by iterative substitution:

a (2) = a (1) + 21

a (3) = a (2) + 21 = (a (1) + 21) + 21 = a (1) + 2×21

a (4) = a (3) + 21 = (a (1) + 2×21) + 21 = a (1) + 3×21

and so on, with the general pattern

a (n) = a (1) + 21 (n - 1) = 21n - 20

Now, we're told that the sum of some number N of terms in this sequence is 2332. In other words, the N-th partial sum of the sequence is

a (1) + a (2) + a (3) + … + a (N - 1) + a (N) = 2332

or more compactly,

$\displaystyle\sum_{n=1}^N a(n) = 2332$

It's important to note that N must be some positive integer.

Replace a (n) by the explicit rule:

$\displaystyle\sum_{n=1}^N (21n-20) = 2332$

Expand the sum on the left as

$\displaystyle 21 \sum_{n=1}^N n-20\sum_{n=1}^N1 = 2332$

and recall the formulas,

$\displaystyle\sum_{k=1}^n1=\underbrace{1+1+\cdots+1}_{n\text{ times}}=n$

$\displaystyle\sum_{k=1}^nk=1+2+3+\cdots+n=\frac{n(n+1)}2$

So the sum of the first N terms of a (n) is such that

21 × N (N + 1)/2 - 20N = 2332

Solve for N :

21 (N ² + N) - 40N = 4664

21 N ² - 19 N - 4664 = 0

Now for the problem I mentioned at the start: this polynomial has no rational roots, and instead

N = (19 ± √392,137)/42 ≈ -14.45 or 15.36

so there is no positive integer N for which the first N terms of the sum add up to 2332.

4 0

2 years ago

1/2 - 20 [ 14- 3 4/15] / 6/5 ^2

lidiya [134]

Step-by-step explanation:

hope it's helpful for you

6 0

3 years ago