All categories

2 years ago

There are 240 calories in 4 ounces of honey. How many calories are in 1 pound? (Note: 1 pound = 16 ounces)

Mathematics

2 answers:

Amiraneli [1.4K]2 years ago

6 0

Answer:960

Step-by-step explanation:

4 Ounces=240 Calories

16 Ounces / 4 Ounces = 4

MAIN ANSWER.!! 4 x 240 = 960 Ounces.

kolezko [41]2 years ago

6 0

Answer:960

Step-by-step explanation:

4 Ounces=240 Calories

16 Ounces / 4 Ounces = 4

4 x 240 = 960 Ounces.

You might be interested in

Simplify 0.75 (5v+11)-0.25 (11+v)

wariber [46]

<span>2 • (7v + 11) over 4 is your answer :)</span>

5 0

3 years ago

Merle opened a new savings account she deposited 40,000 at 10% compounded semiannually at the start of the fourth year merle dep

UNO [17]

happy first question!!!!

4 0

2 years ago

You stuffed 108 envelopes in 45 minutes. At this rate, how many envelopes can you stuff in 2 hours?

Anuta_ua [19.1K]

Answer:

288

Step-by-step explanation:

8 0

3 years ago

The lines below are parallel. If the slope of the green line is -2, what is the

Andreas93 [3]

Answer:

C. -2

Step-by-step explanation:

All parallel lines have the same slope

4 0

2 years ago

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

3 years ago