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1 year ago

How do we know that we Enlarged by 2?

Mathematics

2 answers:

vitfil [10]1 year ago

8 0

Answer:

ggg

Step-by-step explanation:

sweet-ann [11.9K]1 year ago

6 0

I think you multiply everything by 2 but I’m not fully sure

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At Westside Middle School, 10 percent, or 140 students, participate in soccer. How many students are in the school? Check all th

Alexeev081 [22]

Answer:

140 + 10 = 150, so 150 + 140 = 290 students. There are 140 + 10 = 150 students at Westside Middle School.

Step-by-step explanation:

5 0

2 years ago

Decrease £120 by 45%

Dovator [93]

120 * 45 % = 120 *0,45 = 54 £

120 -54 = 66 £

7 0

2 years ago

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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

If f(x) = – x2 + 6x - 1 and g(x) = 3x<br> - 1, find (f + g)(x).

lesantik [10]

Answer:

-x^2 +9x -2

Step-by-step explanation:

f(x) = – x^2 + 6x - 1 and g(x) = 3x - 1,

(f + g)(x) = f(x) + g(x)

= – x^2 + 6x - 1 + 3x - 1

Combine like terms

= -x^2 +9x -2

6 0

2 years ago

What is the product of the complex numbers (-3i+4) and (3i+4)?

Nataly_w [17]

Answer: 25

Step-by-step explanation:

Finding the product of complex numbers is not different to finding the product of real numbers

(-3i+4) * (3i+4) = -9i^2 + 12i - 12i + 16 = 25

5 0

2 years ago

Read 2 more answers