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2 years ago

What are the dimensions of the matrix?

Mathematics

1 answer:

emmasim [6.3K]2 years ago

3 0

Answer:

4 x 3

Step-by-step explanation:

The order of a matrix is given by stating the number of rows followed by the number of columns.

The matrix here has 4 rows and 3 columns thus order is

4 x 3

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What is the measure in each angle? A? B? C?

andrey2020 [161]

Answer:

∠A = 48°, ∠B = 48°, and ∠C = 84°

Step-by-step explanation:

AB = CB Given

m ∠A = m ∠C If opposite sides are =, then those opposite angles are =

6(x - 3) = 4(x + 1)

6x - 18 = 4x + 4

2x - 18 = 4

2x = 22

x = 11

m ∠A = 6(11 - 3) = 6(8) = 48°

m ∠C = 4(11+ 1) = 4(12) = 48°

The sum of the angles in a triangle = 180°

48° + 48° m ∠B = 180°

96°+ m ∠B = 180°

84° = m ∠B

5 0

1 year ago

Find the output for each input of y = -2x.

Bezzdna [24]

Its the first one because -3(-2)=6, 1(-2)=-2, 2(-2)=4 are all correct

3 0

2 years ago

Read 2 more answers

[PLEASE HELP] Each of these statements describe a transformation of a graph of y = x, The which of the statements correctly desc

Sphinxa [80]

Answer:

Step-by-step explanation:

Adding the 7 to the input (x) will increase the output (y) by 7. Therefore, the graph is translated 7 units up.

5 0

2 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

2 years ago

The last choice is “NEITHER”

VARVARA [1.3K]

The answer to this problem is neither. To find this you would just plug in each ordered pair into the equation for the x and y values. Then you would compare if they are equal

5 0

2 years ago