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3 years ago

(4.8). (-8.16) find the slope of the line through each pair of points.

Mathematics

2 answers:

slamgirl [31]3 years ago

5 0

Answer:

16-8/-8-4 = 8/-16 = 1/-2

dalvyx [7]3 years ago

4 0

Answer:

- $\frac{3}{4}$

Step-by-step explanation:

$\frac{16-8}{-8-4}$

- $\frac{3}{4}$

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What is the expanded form of 232÷4

Annette [7]

(2x100) (3x10) (2x1) / (4x1)

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3 years ago

In each of these situations light refracts, except for one. Identify the case in which light is NOT refracted.

Eva8 [605]

B) sunlight travels through space - there is nothing to bounce off of until it hits an atmophere

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2 years ago

Read 2 more answers

VERY IMPORTANT... write the equation of the line that has a y intercept of (0.4) and a slope of -3/2

shutvik [7]

Answer:

y = -3/2 x + 4

Step-by-step explanation:

y-4 = -3/2 ( x-0)

y = -3/2 x + 4

7 0

2 years ago

The length of a rectangular garden ABCD is 9 feet more than its width. It is surrounded by a brick walkway 4 feet wide as shown

Anna71 [15]

Answer:

16.5 ft by 25.5 ft

Step-by-step explanation:

Let w represent the width of the garden in feet. Then w+9 is the garden's length, and w(w+9) represents its area.

The surrounding walkway adds 8 feet to each dimension, so the total area of the garden with the walkway is ...

(w+8)(w+9+8) = w^2 +25w +136

If we subtract the area of the garden itself, then the remaining area is that of the walkway:

(w^2 +25w +136) - (w(w+9)) = 400

16w + 136 = 400 . . .simplify

16w = 264 . . . . . . . . subtract 136

264/16 = w = 16.5 . . . . . width of the garden in feet

w+9 = 25.5 . . . . . . . . . . .length of the garden in feet

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

2 years ago