Ask question

Ask question

All categories

Lady bird [3.3K]

3 years ago

10

james has a card collection with 64 basketball cards, 32 football cards, and 24 baseball cards. he wants to arrange the cards in

equal piles, with only one type of card in each pile. How many can he put in each pile?

1 answer:

____ [38]3 years ago

5 0

64+32=96
96+24=120
120÷3=40
Lee can put 40 cards in each pile.
Your final answer is 40.

You might be interested in

A light shines from the top of a pole 50ft high. a ball is dropped from the same height from a point 30 ft away from the light.

belka [17]

The speed of the ball is
ds/dt = 32t
At t =1/2 s
ds/dt = 16 ft/s
The distance from the ground
50 - 16(1/2)^2 = 46 ft
The triangles formed are similar
50/46 = (30 + x)/x
x = 345 ft

50 / (50 - s) = (30 + x)/x
Taking the derivative and substituing
ds/dt = 16
and
Solve for dx/dt

7 0

3 years ago

Are 45/75 and 90/150 proportional?

lesya [120]

Yes they are
. 90/150= 3/5
so 45/75 and 3/5.

3 0

3 years ago

Read 2 more answers

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:

<u>Taylor series</u> 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}}$

$\text{f}(x)=e^{4x} \implies \text{f}(1)=e^4$

$\text{f}\:'(x)=4e^{4x} \implies \text{f}\:'(1)=4e^4$

$\text{f}\:''(x)=16e^{4x} \implies \text{f}\:''(1)=16e^4$

Substituting the values in the series expansion gives:

$e^{4x}=e^4+4e^4(x-1)+\dfrac{16e^4}{2}(x-1)^2+...$

Factoring out e⁴:

$e^{4x}=e^4\left[1+4(x-1)+8}(x-1)^2+...\right]$

<u>Taylor Series summation notation</u>:

$\displaystyle \text{f}(x)=\sum^{\infty}_{n=0} \dfrac{\text{f}\:^{(n)}(a)}{n!}(x-a)^n$

Therefore:

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

7 0

1 year ago

Trigonometric Ratios

Burka [1]

Answer:

Sis I can't see

Step-by-step explanation:

.

.

.

.

.

Bnnn t is a great place to

7 0

3 years ago

The equation of the graphed line is 2x – y = –6.<br><br><br> What is the x-intercept of the graph?

Bas_tet [7]

U can either look at ur graph...and where the line crosses the x axis is ur x intercept...which is (-3,0)

OR

take ur equation and sub in 0 for y, and solve for x, ur x intercept.
2x - y = -6
2x - 0 = -6
2x = -6
x = -6/2
x = -3.....so ur x intercept is (-3,0)

4 0

2 years ago

Read 2 more answers