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katovenus [111]
3 years ago
15

What is the standard form of 0.004​

Mathematics
1 answer:
Andreas93 [3]3 years ago
3 0

Answer:

4 * 10^-3

Step-by-step explanation:

10^-3 = 0.001

4 * 10^-3 = 0.004

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Darian gets paid $6.35 an hour to help out at his dad's shop. He also mows his neighbor's lawn each week for $25. He earned $119
nevsk [136]
$119.85 is the total amount he earned in 2 weeks
He mows his neighbor's lawn each week for $25...So if he mows for 2 weeks it should be
$25×2=$50
Darian gets paid $6.35 an hour to help out at his dad's shop...
Let X be the number of hours Darian worked at his dad's shop in the two weeks
So we can now make an equation to find X

(X×$6.35)+($25×2)=$119.85
(X×$6.35)+$50=$119.85
X×$6.35=$119.85-$50
X×$6.35=$69.85
X=$69.85÷$6.35
X=11

Therefore the number of hours Darian worked at his dad's shop in those two weeks is 11
4 0
3 years ago
Read 2 more answers
Which number line represents the solution set for the inequality {X24?
zhannawk [14.2K]

Answer:

A.) will be the correct answer to your problem

6 0
4 years ago
Can you help me for number 36, please thank you!
EastWind [94]
The slope is 20/13. Yes, the slope is steeper because without the lip on the slide you would travel downward faster.
6 0
3 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
I need some help on my exit ticket lol
Nina [5.8K]

Answer:

The area of a circle can be found using the formula A=\pi r^2

Step-by-step explanation:

8 0
3 years ago
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