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3 years ago

I’ll give brainlist to correct answers

Mathematics

2 answers:

hoa [83]3 years ago

3 0

Answer:stop stealing points

Step-by-step explanation:

maxonik [38]3 years ago

3 0

Answer:

a) 1

b) 7

c) -3

Step-by-step explanation:

"Adding" and "up" indicate a positive integer.

"Down" indicates a negative integer.

I'm going to assume that a) refers to one plate.

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20 points!

meriva

All circles are similar because all circle have same shape.

However Two circles are congruent , if and only if their radii are congruent.

In our question we are asked when are all circles similar given if their radii are congruent.

So we can say this statement is false because no matter the radii are congruent or not , two or more circles are always similar because of their same shape no matter what the measure of their radii is. Only if we are asked when they are congruent, we will consider the radii part.

Answer is false.

4 0

3 years ago

Read 2 more answers

Need some help with multiplying algebraic fractions

Nezavi [6.7K]

Answer:

a=10, b=5, c=2, d=12

Step-by-step explanation:

Multiply each variable together

5*2=10

8*3=24

10 $x^{3}$ $y^{2}$ /24x

the coefficients (numbers) can each be divided by 2

5 $x^{3} y^{2}$ /12x

subtract the "x" exponents together

5 $x^{2} y^{2}$ /12

8 0

3 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}$

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