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2 years ago

A candy box measures 8.2 cm in diameter and 15 cm in height. How many 2 cm cubed pieces of candy can fit into the box?

1 answer:

8 0

60 2cm candy prices can fit because if you multiply 8.2 by 15 it equals a decimal and if you divide it by 2 that the nearest whole number you’ll get is 60, so the answer has to be 60.

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1)A fish tank has a volume of 520 in. it is emptied at a constant rate of 10 cubic inches every minute. Which equation represent

disa [49]

Answer:

v=520 -10t

Step-by-step explanation:

the rate of decrease is given a variable to calculate the rate of change. v is represented by 520 - the number of minutes.

4 0

2 years ago

Solve for m. –8m = 3 − 9m m =

Oksanka [162]

Answer:

m=3

Step-by-step explanation:

–8m = 3 − 9m

-8+9m = 3

m = 3

8 0

2 years ago

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HELP PLS I WILL MARK BRAINLYIiEST ASAP!!!!

kicyunya [14]

Answer: 50.84%

Step-by-step explanation:

(89-59) /59 = 0.5084/100 = 50.84%

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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

2 years ago

Suppose △DFC≅△SNP . What is the corresponding congruent part for each segment or angle? Drag the answers into the boxes to match

timurjin [86]

<FCD

<SNP

Look at the vertex of each angle and match it to the other triangle (tri. DFC is congruent to tri. SNP) F=N, which is the vertex, D=S, C=P, DF=SN, and so on

7 0

3 years ago

Read 2 more answers