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3 years ago

Which equation can be used to solve for in the following diagram?

Mathematics

1 answer:

Ivahew [28]3 years ago

5 0

Answer:

the Answer is 3x+12)+X=180

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Help please !!!!!!!!

AlekseyPX

Answer:

Place the left of the dot on -5 and the right one on +7

That should do it :)

5 0

3 years ago

Collin did the work to see if 10 is a solution to the equation r/4=2.5

Natasha2012 [34]

Answer:

Yes, because if you substitute 10 for r in the equation and simplify, you find that the equation is true.

Step-by-step explanation:

$\frac{r}{4} = 2.5 \\ lhs = \frac{r}{4} \\ = \frac{10}{4} \\ = 2.5 \\ = rhs \\ \therefore \: \frac{r}{4} = 2.5 \\$

3 0

3 years ago

40.8 gallons of paint among 8 containers how much paint is in each container

LenKa [72]

5.1 gallons of paint are in each container.

4 0

3 years ago

Read 2 more answers

Select the correct answer. Why is it important to look at your audience when speaking? A. So that they don't get bored B. So tha

Norma-Jean [14]

Answer:

Step-by-step explanation:

so you can get the attention and intrest them in your speech

7 0

3 years ago

Read 2 more answers

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