All categories

3 years ago

I NEED HELP ASAP!!!!!

Mathematics

2 answers:

crimeas [40]3 years ago

8 0

Answer:

Skewed to the right

Step-by-step explanation:

i bet your in 6 or 7 grade

Crazy boy [7]3 years ago

4 0

Answer:

skewed right

Step-by-step explanation:

You might be interested in

Can someone help me with this (will give brainliest) please view the attached image

evablogger [386]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

(a) Use the power series expansions for ex, sin x, cos x, and geometric series to find the first three nonzero terms in the powe

Fofino [41]

Answer:

a) $\mathbf{4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} ...}$

b) See Below for proper explanation

Step-by-step explanation:

a) The objective here is to Use the power series expansions for ex, sin x, cos x, and geometric series to find the first three nonzero terms in the power series expansion of the given function.

The function is $e^x + 3 \ cos \ x$

The expansion is of $e^x$ is $e^x = 1 + \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + ...$

The expansion of cos x is $cos \ x = 1 - \dfrac{x^2}{2!}+ \dfrac{x^4}{4!}- \dfrac{x^6}{6!}+ ...$

Therefore; $e^x + 3 \ cos \ x = 1 + \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + ... 3[1 - \dfrac{x^2}{2!}+ \dfrac{x^4}{4!}- \dfrac{x^6}{6!}+ ...]$

$e^x + 3 \ cos \ x = 4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} + \dfrac{x^3}{3!}+ ...$

Thus, the first three terms of the above series are:

$\mathbf{4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} ...}$

The series for $e^x + 3 \ cos \ x$ is $\sum \limits^{\infty}_{x=0} \dfrac{x^x}{n!} + 3 \sum \limits^{\infty}_{x=0} ( -1 )^x \dfrac{x^{2x}}{(2n)!}$

let consider the series; $\sum \limits^{\infty}_{x=0} \dfrac{x^x}{n!}$

$|\frac{a_x+1}{a_x}| = | \frac{x^{n+1}}{(n+1)!} * \frac{n!}{x^x}| = |\frac{x}{(n+1)}| \to 0 \ as \ n \to \infty$

Thus it converges for all value of x

Let also consider the series $\sum \limits^{\infty}_{x=0}(-1)^x\dfrac{x^{2n}}{(2n)!}$

It also converges for all values of x

7 0

3 years ago

The sum of three consecutive even integers is 36 . find the smallest of them

bonufazy [111]

Answer:

Step-by-step explanation:

You can get the three consecutive integers by using algebraic expressions.

Let's solve.

Let x be the smallest even integer.

Let x+2 be the second even integer.

Let x+4 be the largest even integer.

x+x+2+x+4 = 36

3x+6 = 36

3x = 36-6

3x = 30

x = $\frac{30}{3}$

x = 10

Therefore, 10 is the smallest even integer.

x+2 = 10+2 = 12

The second even integer is 12.

x+4 = 10+4 = 14

The largest even integer is 14.

Let's check by adding the three even integers.

10+12+14 = 36

36 = 36

It's correct. 

7 0

3 years ago

Solve cos 2m + sin m = 50, 0°< m< 360°

kenny6666 [7]

I tried calculating the problem but couldn't seem to find an answer. I checked on my graphics calculator and on the computer and they provide a "No Solution" answer. The steps shown above lead to a dead end in the calculation.

8 0

3 years ago

How do you solve 8+9

Evgesh-ka [11]

you just add them together to get 17

8 0

3 years ago

Read 2 more answers