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3 years ago

-3(k-8)-(k+5)=23 please explain how to do this

Mathematics

2 answers:

VMariaS [17]3 years ago

6 0

Answer:

k = -1

Step-by-step explanation:

Use the distributive property to eliminate parentheses. That property tells you that the outside factor multiplies each term inside parentheses:

-3(k -8) -(k +5) = 23

-3k -3(-8) -k -(+5) = 23

-3k +24 -k -5 = 23 . . . . simplify

-4k +19 = 23 . . . . . . . . . collect terms

-4k = 4 . . . . . . . . . . . . . subtract 19 from both sides

k = -1 . . . . . . . . . . . . . . . divide both sides by -4

_____

The rule for manipulating equations is, "any operation performed on one side of the equation must also be performed on the other side." Operations include addition, subtraction, multiplication, division, application of a function. Using the distributive property and collecting terms are ways to simplify equations that don't affect the truth of the equal sign (if those are done correctly).

In general, you make use of inverse operations to undo the operations you don't want. The goal is to get the variable by itself on one side of the equal sign.

lora16 [44]3 years ago

3 0

Answer:

k = -1

Step-by-step explanation:

-3(k-8)-(k+5)=23

Distribute

-3k +24 -k-5 = 23

Combine like terms

-4k +19 = 23

Subtract 19 from each side

-4k +19-19 = 23-19

-4k = 4

Divide by -4

-4k/-4 = 4/-4

k = -1

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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 $-4a^3b$ .

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