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3 years ago

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In a box of chocolates the ratio of plain chocolates to milk chocolates is 2:5 what fraction of the chocolates are plain what fr

action of the chocolates are milk

1 answer:

eduard3 years ago

3 0

Answer:

$\frac{2}{7}$ of the chocolates are plain.

$\frac{5}{7}$ of the chocolates are milk.

Step-by-step explanation:

The ratio of plain chocolates to milk chocolates is 2:5

This means that for each 2+5 = 7 chocolates, 2 are plain and 5 are milk.

So $\frac{2}{7}$ of the chocolates are plain.

And $\frac{5}{7}$ of the chocolates are milk.

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Help plz I've been stuck

stepladder [879]

Answer:

The answer is D. $(\frac{1}{5} )^{2}$

Step-by-step explanation:

$5^{4} *(5^{-3} )^{2}$

simplify the expression by multiplying

$5^{4} *5^{-6}$

calculate the product

$5^{-2}$

express with a positive exponent using $a^{-n} =\frac{1}{a^{n} }$

$\frac{1}{5^{2} }$

6 0

3 years ago

Read 2 more answers

A) 7.7 ft<br> b) 13.4 ft<br> c) 10.3 ft<br> d) 12.3 ft

ELEN [110]

Answer:

D. 12.3 ft

Step-by-step explanation:

This problem requires the use of trigonometric ratios. This one specifically uses the cosine ratio as it provides the hypotenuse and is asking for the side that is adjacent to the angle.

cos(40°)=a/16

cos(40°)×16=a/16×16

cos(40°)×16=a

a=12.256 ft

The length of side a is D. 12.3 ft.

3 0

4 years ago

PLEASE HURRY I NEED MY WORK TO BE SHOWN!!!!! 15 POINTS!!! CORRECT ANSWERS ONLY PLEASE!!!

Harlamova29_29 [7]

Answer:

f(-4) = 72, f(x + 5) = 3x² + 23x + 36

Step-by-step explanation:

f(-4) = 3(-4)² - 7(-4) - 4

= 48 - (-28) - 4

f(-4) = 72

f(x + 5) = 3(x + 5)² - 7(x + 5) - 4

= 3(x + 5)(x + 5) - 7(x + 5) - 4

= 3(x² + 10x + 25) - 7(x + 5) - 4

= 3x² + 30x + 75 - 7x - 35 - 4

f(x + 5) = 3x² + 23x + 36

4 0

3 years ago

Use the definition of Taylor series to find the Taylor series, centered at c, for the function. f(x) = sin x, c = 3π/4

anyanavicka [17]

Answer:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Step-by-step explanation:

Given

$f(x) = \sin x\\$

$c = \frac{3\pi}{4}$

Required

Find the Taylor series

The Taylor series of a function is defines as:

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

We have:

$c = \frac{3\pi}{4}$

$f(x) = \sin x\\$

$f(c) = \sin(c)$

$f(c) = \sin(\frac{3\pi}{4})$

This gives:

$f(c) = \frac{1}{\sqrt 2}$

We have:

$f(c) = \sin(\frac{3\pi}{4})$

Differentiate

$f'(c) = \cos(\frac{3\pi}{4})$

This gives:

$f'(c) = -\frac{1}{\sqrt 2}$

We have:

$f'(c) = \cos(\frac{3\pi}{4})$

Differentiate

$f"(c) = -\sin(\frac{3\pi}{4})$

This gives:

$f"(c) = -\frac{1}{\sqrt 2}$

We have:

$f"(c) = -\sin(\frac{3\pi}{4})$

Differentiate

$f"'(c) = -\cos(\frac{3\pi}{4})$

This gives:

$f"'(c) = - * -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

So, we have:

$f(c) = \frac{1}{\sqrt 2}$

$f'(c) = -\frac{1}{\sqrt 2}$

$f"(c) = -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

becomes

$f(x) = \frac{1}{\sqrt 2} - \frac{1}{\sqrt 2}(x - \frac{3\pi}{4}) -\frac{1/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{1/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Rewrite as:

$f(x) = \frac{1}{\sqrt 2} + \frac{(-1)}{\sqrt 2}(x - \frac{3\pi}{4}) +\frac{(-1)/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{(-1)^2/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Generally, the expression becomes

$f(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Hence:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

3 0

3 years ago

In the first half of a basketball game, a player scored 9 points on free throws and then scored a number of 2-point shots. In th

KatRina [158]

This question is incomplete

Complete Question

In the first half of a basketball game, a player scored 9 points on free throws and then scored a number of 2-point shots. In the second half, the player scored the same number of 3-point shots as the number of 2-point shots scored in the first half. Which expression represents the total number of points the player scored in the game?

a) 2x + 3x + 9

b) 2x + 3 + 9

c) 2x + 3x + 9x

d) 2 + 3x + 9

Answer:

a) 2x + 3x + 9

Step-by-step explanation:

Let the number of points shots a player scores = x

In a free throw, the player scored 9 points = 9

The player also scored a number of 2-point shots = 2x

In the second half, the player scored the same number of 3-point shots as the number of 2-point shots scored in the first half = 3x

The expression represents the total number of points the player scored in the game =

2x + 3x + 9

4 0

3 years ago