All categories

3 years ago

The prime number which is also even is a) 0 b) -2 c) 2 d) 6

Mathematics

2 answers:

Archy [21]3 years ago

7 0

Answer:

c) 2

Step-by-step explanation:

a prime number is a number which is only completely divisible by 1 and itself

lianna [129]3 years ago

4 0

Answer:

the answer is 2

good luck bro

Step-by-step explanation:

C) 2

You might be interested in

X+2=2(2x-14) solve for x.

LekaFEV [45]

X =10.

First distribute the 2(2x-14) and get 4x-28.

x+2-4x-28. Subtract x from each side to get 2=3x-28.

Add 28 to both sides to equal 30=3x. Divide each side by 3 to get your final Answer, x=10.

3 0

3 years ago

0.000701 to a scientific notation

slega [8]

7.01×10^-4. udufhfhf

7 0

3 years ago

Read 2 more answers

translate the sentence into an equation. the product of 8 and a number increased by 4 is 60.a. 8y 4 = 60b. 4y 8 = 56c. 8 ÷ y 4 =

krok68 [10]

Hello,

"T<span>he product of 8 and a number increased by 4" may be understood by 2 ways:
1) it is the number which is increased
2) it is the product which is increased.

1) 8*(y+4) =8y+32

2) (8y)+4=8y+4

Seeing answers
:answer A: 8y+4

</span>

6 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}$

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

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

3 0

3 years ago

Solve the system by using a matrix equation (Picture provided)

Katyanochek1 [597]

Answer:

Option b is correct (8,13).

Step-by-step explanation:

7x - 4y = 4

10x - 6y =2

it can be represented in matrix form as $\left[\begin{array}{cc}7&-4\\10&-6\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}4\\2\end{array}\right]$