All categories

3 years ago

Simplify -4b + (-9k) -6 - 3b + 12

Mathematics

1 answer:

Olenka [21]3 years ago

6 0

Answer:

-7b-9k+6

Step-by-step explanation:

-4b+(-9k)-6-3b+12

(-4b-3b)+(-9k)+(-6+12)

-7b+(-9k)+6

-7b-9k+6

You might be interested in

(9x²+24x-8)-(5x²+3x+10)

motikmotik

Answer:

4x^2 + 21x - 18

Step-by-step explanation:

(9x^2+24x-8)-(5x^2+3x+10)

9x^2+24x-8-5x^2-3x-10

9x^2 - 5x^2 = 4x^2

24x - 3x = 21x

-8 - 10 = -18

Answer: 4x^2 + 21x - 18

8 0

3 years ago

Write any 4 laws of exponents.<br>

emmainna [20.7K]

$\rule{50}{1}\large\blue\textsf{\textbf{\underline{Question:-}}}\rule{50}{1}$

<em>Write any 4 laws, or properties, of exponents.</em>

<em />

<em> </em> $\rule{50}{1}\large\blue\textsf{\textbf{\underline{Answer and how to solve:-}}}\rule{50}{1}$

$\Large\text{Law number 1:-}$

$\Large\text{$a^n*a^m=a^{n+m}$}$

$\Large\text{It states that:-}$

When we multiply together numbers with the same base,

we add the exponents.

$\Large\text{Law number 2:-}$

$\Large\text{$a^m:a^n=a^{m-n}$}$

$\Large\text{It states that:-}$

When we divide numbers with the same base, we

subtract the exponents.

$\Large\text{Law number 3:-}$

$\Large\text{$(\displaystyle\frac{x}{y}) ^n=\frac{x^n}{y^n}$}$

$\Large\text{It states that:-}$

If we have a fraction to a power, we raise the numerator and

the denominator to that power.

And then last but not least,

$\Large\textit{Law number 4:-}$

$\Large\text{$a^{-m} =\displaystyle\frac{1}{a^m}$}$

$\Large\text{It states that:-}$

If we have a number with a negative exponent, we flip it over.

<h3>Good luck with your studies.</h3>

#TogetherWeGoFar

$\rule{50}{1}\smile\smile\smile\smile\smile\smile\rule{50}{1}$

6 0

2 years ago

Read 2 more answers

If Asher plays each of the remaining seasons for which he is eligible, what is the average number of doubles he needs to hit per

PtichkaEL [24]

Answer:

the

Step-by-step explanation:

6 0

3 years ago

Solve the equation15=2y-5

Lesechka [4]

15 = 2Y - 5
20 = 2Y
10 = Y

5 0

3 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago