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3 years ago

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Use properties of operations to complete the equivalent expression. 5(y+9)=_y+_ 5(y+9)=nothingy+nothing

1 answer:

laiz [17]3 years ago

5 0

Answer:

5y+45

Step-by-step explanation:

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Find the solutions to the equation below.<br> Check all that apply.<br> 4x2 + 4x + 1 = 0

Verizon [17]

Answer:

The solution for the given polynomial $4x^2 + 4x+ 1 = 0$ is $x = (\frac{-1}{2})$

Step-by-step explanation:

Here, the given quadratic equation is : $4x^2 + 4x+ 1 = 0$

Now, here solving the given equation by SPLITTING THE MIDDLE TERM:

$4x^2 + 4x+ 1 = 0 = 4x^2 + 2x+ 2x + 1 = 0\\\implies 2x( 2x + 1) + 1(2x + 1) = 0\\\implies (2x +1)(2x +1) = 0\\$

⇒ either (2x+1) = 0, or ( 2x + 1) = 0

⇒ x = -1/2 or x = -1/2

Here, in both the cases, the value of x is equal i.e x = -1/2

So, the given function has TWO IDENTICAL ROOTS.

Hence the solution for the given polynomial $4x^2 + 4x+ 1 = 0$ is $x = (\frac{-1}{2})$

4 0

3 years ago

The table shows the approximate area in square miles of the largest and smallest states in the United States.

alexandr402 [8]

Who’s tryna hop on cod tho

8 0

2 years ago

PLZ HELP ME !!!!!!!!!!!!!!<br> 2 questions

disa [49]

Answer: D

Step-by-step explanation: I think

6 0

2 years ago

I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).

Aleksandr-060686 [28]

Answer:

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

Step-by-step explanation:

$\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$

Let's first isolate the trig function.

Add 1 one on both sides:

$\sqrt{3} \tan(x-\frac{\pi}{8})=1$

Divide both sides by $\sqrt{3}$ :

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

Now recall $\tan(u)=\frac{\sin(u)}{\cos(u)}$ .

$\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

or

$\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$

The first ratio I have can be found using $\frac{\pi}{6}$ in the first rotation of the unit circle.

The second ratio I have can be found using $\frac{7\pi}{6}$ you can see this is on the same line as the $\frac{\pi}{6}$ so you could write $\frac{7\pi}{6}$ as $\frac{\pi}{6}+\pi$ .

So this means the following:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

is true when $x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

where $n$ is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

$x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

Add $\frac{\pi}{8}$ on both sides:

$x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi$

Find common denominator between the first two terms on the right.

That is 24.

$x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi$

$x=\frac{7\pi}{24}+n \pi$ (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval $[0,2\pi)$ .

So if $\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$ and we let $u=x-\frac{\pi}{8}$ , then solving for $x$ gives us:

$u+\frac{\pi}{8}=x$ ( I just added $\frac{\pi}{8}$ on both sides.)

So recall $0\le x$ .

Then $0 \le u+\frac{\pi}{8}$ .

Subtract $\frac{\pi}{8}$ on both sides:

$-\frac{\pi}{8}\le u$

Simplify:

$-\frac{\pi}{8}\le u$

$-\frac{\pi}{8}\le u$

So we want to find solutions to:

$\tan(u)=\frac{1}{\sqrt{3}}$ with the condition:

$-\frac{\pi}{8}\le u$

That's just at $\frac{\pi}{6}$ and $\frac{7\pi}{6}$

So now adding $\frac{\pi}{8}$ to both gives us the solutions to:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$ in the interval:

$0\le x$ .

The solutions we are looking for are:

$\frac{\pi}{6}+\frac{\pi}{8}$ and $\frac{7\pi}{6}+\frac{\pi}{8}$

Let's simplifying:

$(\frac{1}{6}+\frac{1}{8})\pi$ and $(\frac{7}{6}+\frac{1}{8})\pi$

$\frac{7}{24}\pi$ and $\frac{31}{24}\pi$

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

5 0

3 years ago

Find the nth term of the arithmetic sequences<br> a1=5,d=6,n=11

ki77a [65]

here's the solution,

n = 11
a = 5 ( a = first term )
d = 6 ( d = common difference )

we know,

=》

$nth \: \: term \: = a \: + (n - 1) \times d$

=》

$11th \: \: term = 5 + (11 - 1) \times 6$

=》

$11th \: \: term = 5 + (10 \times 6)$

=》

$11th \: \: term = 5 + 60$

=》

$11th \: \: term = 65$

nth term ( 11th term ) = 65

3 0

3 years ago