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2 years ago

13

Please answer this correctly correctly without making mistakes I want ace expert and genius people to answer this correctly with

out making mistakes

2 answers:

brilliants [131]2 years ago

5 0

Answer: 6

10 + -2 = 8

8 + -2 = 6

sergejj [24]2 years ago

4 0

I think the answer is 12

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You ran into Pedro's for a quick lunch and spent

SCORPION-xisa [38]

Answer:

The total answer is $14.16 because the tip would be $2.16 so add it with 12 and you get $14.16.

6 0

3 years ago

For every 8 red pens on the teachers desk there are 12 pencils. If she has 60 writing implements, does she have enough red pens

gavmur [86]

Answer:No she does not

Step-by-step explanation:

8pens + 12pecils= 20writing inplements

20+8+12=40

20+8+12=60

You needed 3 sets of 8 pens and 3 sets of 12 pencils to get 60.

3*8=24

24 is less than 60, so she does not have enough pens for 60 students

7 0

2 years ago

What is fourteen times one thousand ?*

ioda

The trick with answering your 10s, 100s, 1000s, etc, is just adding zeroes onto the ends of the first number. So, since this is 14 times 1000, you would add 3 zeroes onto 14. So your answer is 14,000. Hope this helped!

7 0

3 years ago

Read 2 more answers

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

3 years ago

A local hamburger shop sold a combined total of 685 hamburgers and cheeseburgers on Wednesday. There were 65 fewer cheeseburgers

noname [10]

Answer:

620

Step-by-step explanation:

I might be wrong.

But based on the question both burgers are sold on a Wednesday so the question is asking how many cheeseburgers were sold on a Wednesday (which is the same day they both were sold) so I think you subtract.

so 685-65.

8 0

3 years ago