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3 years ago

Plz help me with this and u can change the slide too

Mathematics

1 answer:

Nadusha1986 [10]3 years ago

4 0

The 1 slide it is intersecting the second slide is the 4x-y=5 slide 3 is the same and 4th slide is different

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How many real roots does the function

icang [17]

Answer:

Option C is correct i.e. 2.

Step-by-step explanation:

Given the function is f(x) = x² +8x -2.

We can compare it with general quadratic expression i.e. ax² +bx +c.

Then a = 1, b = 8, c = -2.

We can find the number of real root by finding discriminant of the equation ax² +bx +c =0 as follows:-

D = b² -4ac

D = 8² -4*1*-2

D = 64 +8

D = 72.

When D is a positive value, then we have two real roots of the equation.

Hence, option C is correct i.e. 2.

6 0

3 years ago

Solutions to x^=-11x+4

anastassius [24]

x=-11x+4

x+11x=-11x+4+11x

12x=4

x= 4/12

<h2>Answer: x=1/3</h2>

5 0

4 years ago

WILL MARK BRAINLIEST IF CORRECT! ANSWER A, AND B, I GOT 5 MINS

Schach [20]

Answer:

dollar amount in decimal is 0.55

in fraction form is 11/20

8 0

3 years ago

If 81 is added to a number the result is 39 less than three times the number

Deffense [45]

let x =the number

Step-by-step explanation:

81+x=3x-39

group like terms

81+39=3x-x

120=2x

120/2=2x/2

60=x

therefore x =60

4 0

3 years ago

Read 2 more answers

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

3 years ago