All categories

3 years ago

PLEASE HELPPPP I WILL MARK YOU AS BRAINLIEST!!!!!

Mathematics

1 answer:

statuscvo [17]3 years ago

6 0

Answer:

Let a be the cost of each apple.

Let p be the cost of each pear.

Q1

Cost of 12 apples = 12a

Q2

Cost of 4 pears = 4p

Q3

TOTAL COST = 12a + 4p

You might be interested in

PLEASE HELP THESE ARE MY ONLY POINT ;-; PLEASE

Zolol [24]

Answer:

a) max 12000 , min 0

b) no solution

c) pricing based on profit

Step-by-step explanation:

a) look at the parabola

b) there is more than one x (0)

c) the y intercepts represent his profit based on how much he priced the items at

3 0

2 years ago

7/5=10.5/x Help plz

Natasha_Volkova [10]

7.5. 10.5*5 and divide that by 7

6 0

3 years ago

Read 2 more answers

Which is 2/3 of 18 A. 27 B. 8 C. 12 D. 6

ahrayia [7]

Answer:

C. 12

Step-by-step explanation:

We know you have the number 18 and the fraction 2/3. The problem is we want to find out what 2/3 of 18 is. Well, if 18 divided by the denominator of that fraction (3) is 6 than we just add 6 because the question is asking for 2/3 NOT 1/3.

6 0

3 years ago

Consider the function f(x)equalscosine left parenthesis x squared right parenthesis. a. Differentiate the Taylor series about 0

dybincka [34]

I suppose you mean

$f(x)=\cos(x^2)$

Recall that

$\cos x=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}$

which converges everywhere. Then by substitution,

$\cos(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{(x^2)^{2n}}{(2n)!}=\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n)!}$

which also converges everywhere (and we can confirm this via the ratio test, for instance).

a. Differentiating the Taylor series gives

$f'(x)=\displaystyle4\sum_{n=1}^\infty(-1)^n\frac{nx^{4n-1}}{(2n)!}$

(starting at $n=1$ because the summand is 0 when $n=0$ )

b. Naturally, the differentiated series represents

$f'(x)=-2x\sin(x^2)$

To see this, recalling the series for $\sin x$ , we know

$\sin(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^{n-1}\frac{x^{4n+2}}{(2n+1)!}$

Multiplying by $-2x$ gives

$-x\sin(x^2)=\displaystyle2x\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n+1)!}$

and from here,

$-2x\sin(x^2)=\displaystyle 2x\sum_{n=0}^\infty(-1)^n\frac{2nx^{4n}}{(2n)(2n+1)!}$

$-2x\sin(x^2)=\displaystyle 4x\sum_{n=0}^\infty(-1)^n\frac{nx^{4n}}{(2n)!}=f'(x)$

c. This series also converges everywhere. By the ratio test, the series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(n+1)x^{4(n+1)}}{(2(n+1))!}}{(-1)^n\frac{nx^{4n}}{(2n)!}}\right|=|x|\lim_{n\to\infty}\frac{\frac{n+1}{(2n+2)!}}{\frac n{(2n)!}}=|x|\lim_{n\to\infty}\frac{n+1}{n(2n+2)(2n+1)}$

The limit is 0, so any choice of $x$ satisfies the convergence condition.

3 0

3 years ago

What is one-half the sum of 142 and 8

vichka [17]

75 is one half the sum of 142 and 8

6 0

3 years ago

Read 2 more answers