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3 years ago

Write and equation to represent the following statement 28 is 12 less thank K. solve for K K =

Mathematics

2 answers:

kolezko [41]3 years ago

6 0

Answer:

<h3>Equation : 28 = k - 12</h3>

Step-by-step explanation:

28 is 12 less than k

Let's create an equation:

$28 = k - 12$

Now, let's solve:

$28 = k - 12$

Move variable to L.H.S and change its sign

Similarly, Move constant to R.H.S and change its sign

$- k = - 12 - 28$

Calculate the difference

$- k = - 40$

Change the signs on both sides of the equation

$k = 40$

Hope this helps...

Best regards!!

steposvetlana [31]3 years ago

5 0

Answer:

K = 40

Step-by-step explanation:

As they said that 28 is 12 less than K , it means that you've to add them to get the answer. So , 28 + 12 = 40 which is represented by the variable "K"

Hope it helps and pls mark as brainliest : )

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514 A restaurant's menu features appetizers, 7 entrées, and 3 dessertsA customer orders only oneappetizer, only one entrée, and

yuradex [85]

Simple...there'a a LOT of combinations...

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Just take 5*7=35

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3 years ago

How many 2-digit numbers are multiples of 7 or 5

wolverine [178]

Answer:

Use the prime factorization:

7 * 5 * 1 = 35

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2 years ago

Identify the expression with nonnegative limit values. More info on the pic. PLEASE HELP.

marshall27 [118]

Answer:

$\lim _{x\to 2}\:\frac{x-2}{x^2-2}\\\\ \lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}\\\\ \lim _{x\to \frac{5}{2}}\left\frac{2x^2+x-15}{2x-5}\right$

Step-by-step explanation:

a) $\lim _{x\to 3}\:\frac{x^2-10x+21}{x^2+4x-21}=\lim \:_{x\to \:3}\:\frac{\left(x-7\right)\left(x-3\right)}{\left(x+7\right)\left(x-3\right)}=\lim \:_{x\to \:3}\:\frac{x-7}{x+7}=\frac{3-7}{3+7}=-\frac{4}{10}=-\frac{2}{5}$

b) $\lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=\lim \:_{x\to -\frac{3}{2}}\:\frac{\left(2x+3\right)\left(x-4\right)}{\left(2x+3\right)}=\lim \:\:_{x\to \:-\frac{3}{2}}\:\left(x-4\right)=-\frac{3}{2}-4\\ \\ \lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=-\frac{11}{2}$

c) $\lim _{x\to 2}\:\frac{x-2}{x^2-2}=\frac{2-2}{\left(2\right)^2-2}=\frac{0}{4-2}=0$

d) $\lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}=\lim _{x\to 11}\:\frac{\left(x-11\right)\left(x+17\right)}{\left(x-11\right)\left(x+14\right)}=\lim _{x\to 11}\:\frac{\left(x+17\right)}{\left(x+14\right)}=\frac{11+17}{11+14}=\frac{28}{25}$

e) $\lim _{x\to 3}\:\frac{x^2-8x+15}{x-3}=\lim \:_{x\to \:3}\:\frac{\left(x-3\right)\left(x-5\right)}{x-3}=\lim _{x\to 3}\left(x-5\right)=3-5=-2$

f) $\lim _{x\to \frac{5}{2}}\left(\frac{2x^2+x-15}{2x-5}\right)=\lim \:_{x\to \:\frac{5}{2}}\frac{\left(2x-5\right)\left(x+3\right)}{2x-5}=\lim \:\:_{x\to \:\:\frac{5}{2}}\left(x+3\right)=\frac{5}{2}+3=\frac{11}{2}$