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3 years ago

Help please stuck again .

Mathematics

1 answer:

xenn [34]3 years ago

3 0

If the polygons are similar, then the top side is equal to one half of the left side.
Since side (x -1) = 8, then x = 9.

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Evaluate (-27)^-4/3 ((-27 to the power of -4/3))

masha68 [24]

Answer:

-1/81

Step-by-step explanation:

-27^-4/3

cbrt(-27^-4)

cbrt(-1/531441)

-1/81

5 0

3 years ago

Read 2 more answers

Round 66.42 to the nearest hundredth

Musya8 [376]

Answer:

66.42

Step-by-step explanation:

66.42 is already rounded to the nearest hundredth.

3 0

3 years ago

Complemetary angles are two angles whose measures add up to 90 degrees. The ratio of the measures of two complementary angles is

pochemuha

Let A and B be the two complementary angles.

A = smaller angle = 2x

B = larger angle = 13x

x = some unknown number

Note how the ratio A:B turns into 2x:13x which simplifies to 2:13

A+B = 90 ... because the angles are complementary

2x+13x = 90 ... substitution

15x = 90

x = 90/15

x = 6

A = 2*x = 2*6 = 12 degrees

B = 13*x = 13*6 = 78 degrees

The two angles are 12 degrees and 78 degrees.

Check:

A/B = 12/78 = (2*6)/(13*6) = 2/13, so A:B = 2:13

A+B = 12+78 = 90

3 0

3 years ago

A cube has an edge measuring 4 inches what is the surface area of the cube in inches^2

Arada [10]

Answer:

96^2in

Step-by-step explanation:

4x4=16

16x6=96

5 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