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Ad libitum [116K]
2 years ago
5

PLEASE HELP URGENT NOW PLS

Mathematics
2 answers:
goldfiish [28.3K]2 years ago
8 0

Answer:

<h3>(-3, -2)</h3><h3 />

Step-by-step explanation:

Coordinates of intersection of lines =  x = - 3,   y = - 2

RideAnS [48]2 years ago
5 0

Answer:

  • (-3, -2)

Step-by-step explanation:

Given

  • Graphed lines

To find

  • Coordinates of intersection of lines

As per the graph the intersection point has coordinates:

  • x = - 3, y = - 2

The point is (-3, -2)

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faltersainse [42]

Answer:

dcsdjccjwciwenwenvjlwenfilweknf f

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
How to solve this.im stuck in this question for 20 mins. Plzz answer
Alik [6]

Answer:

Find the minimum or maximum value of the function g (I) = -3x^2 - 6x + 5. Describe the domain and range of the function, and where the function is increasing and decreasing. > -1 all real numbers The function The maximum value is I < 0 The domain is and the range is right of I left -1 is increasing to the of I= and decreasing to the 0 12 :: yo y0 :: 8 :: IS-1 :: -1 :: 0 :: I> 0 :: 1 :: 8 :: < 0 :: left :: all real numbers :: y -8 :: y < 8 :: y8​

Step-by-step explanation:

4 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
Carlos has used 40 cell phone minutes each day for one full week. His cell phone plan allows for a total of 320 minutes per week
Vesnalui [34]
320/40 which you simplify.
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3 years ago
This is another 2 in one so tell me which one is the answer to which question if you only answer one
Dominik [7]

Answer:

14 is 62.5 and 15 is 156

Step-by-step explanation:

5 0
3 years ago
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