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Ahat [919]
3 years ago
14

(15pts) just about area don't have to show work

Mathematics
1 answer:
Kazeer [188]3 years ago
6 0
Good luck with ur quiz I hope u pass
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Eight less than four times a number is less then 56. What are the possible values of that number?
Mashutka [201]

Answer: B. X<12

Step-by-step explanation:

First we need to find out what number multipled by 4 is less than 56

4 x 12 = 48

56-48 = 8

And 4 is lower/less than 12 so that's why it's B

5 0
3 years ago
Read 2 more answers
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
a garden plot is to contain 240 sq. ft. if its length is to be 3 times its width,what should its dimensions be?
SOVA2 [1]
240 ft^2 = L X W
240 = 3W x W
240 = 3W^2
W = 8.94427 ft
L = 3W = 3(8.94427)
L = 26.8328 FT
8 0
3 years ago
Is the following shape a rectangle? How do you know?
pogonyaev
I would think it’s a square
7 0
2 years ago
Drag and drop each expression into the box to correctly classify it as having a positive or negative product
AfilCa [17]

Following expressions will have negative product

(\frac{1}{2})(-\frac{1}{2})\ and\ (-\frac{1}{2})(\frac{1}{2})

Following expressions will have positive product

(-\frac{1}{2})(-\frac{1}{2})\ and\ (\frac{1}{2})(\frac{1}{2})

Further explanation:

We will see at each expression one by one

First expression is:

(-\frac{1}{2})(\frac{1}{2})

In the given expression one term is positive and one term is negative. The product of negative and positive terms is negative.

Second Expression is:

(-\frac{1}{2})(-\frac{1}{2})

Both terms are negative and the product of two negative terms is positive.

Third Expression is:

(\frac{1}{2})(-\frac{1}{2})

The expression has one positive and one negative term so the product will be negative

Fourth Expression is:

(\frac{1}{2})(\frac{1}{2})

Both terms are positive so the product will also be positive

Keywords: Product, Expressions

Learn more about expressions at:

  • brainly.com/question/10699220
  • brainly.com/question/10703930

#LearnwithBrainly

5 0
3 years ago
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