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3 years ago

A box of chocolates costs $9. How much will 15 boxes of chocolate cost? A. $119 b. $120 c. $125 d. $135 e. $145

1 answer:

7 0

To solve this problem, we must multiply the cost of one box of chocolates ($9) by the number of boxes of chocolates purchased (15) to find the total cost of the chocolates.

$9 * 15 = $135

Therefore, your answer is $135, or option D.

Hope this helps!

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Using appreciate property -2/3 × 3/5 + 5/2 - 3/5 × 1/6

o-na [289]

Answer:

The answer for your question is 2.

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Mary is spending money at the average rate of $5 per day. After 14 days she has $68 left. The amount left depends on the number

Radda [10]

Answer:

A. Y-68=-5(x-14)

Step-by-step explanation:

Mary is spending money at the average rate of $5 per day. After 14 days she has $68 left. The amount left depends on the number of days that have passed. write an equation for the situation

We find the equation using the point slope equation of a line. This is given as:

y - y1 = m(x-x1)

The slope intercept form will be

y = -5x + b

since the amount decreases by $5 per day...

Our slope is -5 so m=-5

We have a point (x1,y1) = (14, 68)

Hence, we have:

y-68 = -5(x-14)

Option A is correct

4 0

3 years ago

Find the future value if £7400 is invested for 6 years at 12.8% p.a.

Free_Kalibri [48]

Answer:

5683.2 Euro's

Step-by-step explanation:

p=7400 Euro's

r=12.8*1/100

t=6yrs

I=P*R*T

7400*12.8/100*6

=5683.2 Euro's

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3 years ago

TRUE OR FALSE: FOR ANY FUNCTION, X=F^-1(Y), THEN Y=F(X). PLEASE EXPLAIN

Vesnalui [34]

Answer:

True.

Step-by-step explanation:

For example let f(x) = 3x + 1 then f-1(x) is found as follows

Let f(x) = y = 3x + 1 then

3x = y - 1

x = (y-1) / 3

x = f-1(x) = (x - 1)/3.

So:

Replacing the x in y by (x - 1)/3 :-

x = f(-1)(y) = ( (3x + 1) - 1) / 3

= 3x / 3

= x.

So y = f(x).

7 0

3 years ago

Use the definition of a Taylor series to find the first three non zero terms of the Taylor series for the given function centere

Ket [755]

Answer:

$e^{4x}=e^4+4e^4(x-1)+8e^4(x-1)^2+...$

$\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n$

Step-by-step explanation:

Taylor series expansions of f(x) at the point x = a

$\text{f}(x)=\text{f}(a)+\text{f}\:'(a)(x-a)+\dfrac{\text{f}\:''(a)}{2!}(x-a)^2+\dfrac{\text{f}\:'''(a)}{3!}(x-a)^3+...+\dfrac{\text{f}\:^{(r)}(a)}{r!}(x-a)^r+...$

This expansion is valid only if $\text{f}\:^{(n)}(a)$ exists and is finite for all $n \in \mathbb{N}$ , and for values of x for which the infinite series converges.

$\textsf{Let }\text{f}(x)=e^{4x} \textsf{ and }a=1$

$\text{f}(x)=\text{f}(1)+\text{f}\:'(1)(x-1)+\dfrac{\text{f}\:''(1)}{2!}(x-1)^2+...$

$\boxed{\begin{minipage}{5.5 cm}\underline{Differentiating $e^{f(x)}$}\\\\If $y=e^{f(x)}$, then $\dfrac{\text{d}y}{\text{d}x}=f\:'(x)e^{f(x)}$\\\end{minipage}}$