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2 years ago

Greatest to least of 2, -1, -2, 0, 1

Mathematics

2 answers:

sleet_krkn [62]2 years ago

5 0

Answer:

2,1,0,-1,-2

Step-by-step explanation:

Just find the highest number and start going down.

Wittaler [7]2 years ago

4 0

Answer:

2,1,0,-1,-2

Step-by-step explanation:

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Can anyone help me out please

Brilliant_brown [7]

Answer:

Option (3)

Step-by-step explanation:

We have to divide polynomial given as (x³ - 21x² + 147x - 343) by (x - 7) by synthetic division.

7 | 1 -21 147 -343

↓ 7 -98 343

1 -14 49 0

Therefore, quotient is (x² - 14x + 49) and the remainder will be 0.

Option (3) will be the answer.

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2 years ago

Drag each equation to the value of the n that makes it true

Svetradugi [14.3K]

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2 years ago

Lesson 10.5 practice A Trigonometry

grigory [225]

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3 years ago

Approximate the sum of the convergent infinite series the least usingnumber of terms so that the error is less than .0001. What

____ [38]

Solution :

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ $a_n= \frac{(-1)^n}{n ! 2^n} \ \ \ \ \ a_{n+1}= \frac{(-1)^{n+1}}{(n+1) ! 2^{n+1}}$

Error = $|a_{n+1}|$

Error ≤ 0.0001

$|\frac{(-1)^{n+1}}{(n+1)!2^{n+1}}| \leq 0.0001$

$|\frac{1}{(n+1)!2^{n+1}}| \leq 10^{-4}$

$(n+1)! 2^{n+1} \geq 10000$

Now try, n ≥ 5

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = $\sum_{n=1 }^{5 } \frac{(-1)^n}{n ! 2^n}$ (with error 0.0001)

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = $\frac{-1}{1!2}+\frac{1}{2!2^2}-\frac{1}{3! 2^3}+\frac{1}{4! 2^4}-\frac{1}{5! 2^5}$

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = 0.6065104

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3 years ago

Solve the equation. 40 = x - 20 a-60 b60 c20 d -20

tangare [24]

40 = x - 20

move -20 to the other side

sign changes from -20 to +20

40+20= x-20+20

40+20=x

60=x

Answer:

b 60

4 0

2 years ago