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Nutka1998 [239]
3 years ago
9

Whats the answer to this question and how did you do it?? thanks,,, (:

Mathematics
2 answers:
lutik1710 [3]3 years ago
7 0
5x+3=3x+7\\
5x-3x=7-3\\
2x=4\\
x=2
MrRa [10]3 years ago
5 0
5x+3=3x+7
<em></em>-3     =   -3
2x+3=7
     -3= -3
2x=4
x=2
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I need help please Solve the problems! CHECK PICTURE FOR QUESTION
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Check the picture below.

we know that AL is an angle bisector, so the angle at A gets cuts into two equal halves, we also know the angle at B is 30°, so the triangle ABC is really a 30-60-90 triangle, meaning the angle at A is really a 60° angle, cut in two halves gives us 30° and 30° as you see in the picture.

if the angles at A and B, inside the triangle ABL, are equal, twin angles are only made in an isosceles by twin sides, that means that AL = BL.

Looking at the triangle ALC, we can see is also another 30-60-90 triangle, and we can just use the 30-60-90 rule to get x=CL.

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2 years ago
Please help 10 points asap​
hoa [83]

Answer:

a) 7

b) -8.2

c) 0

d) -7

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f) 121

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Just put the opposite for each one. The absolute value is the distance away from 0.

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ElenaW [278]

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Step-by-step explanation:

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3 years ago
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sergejj [24]

Answer:

A

Step-by-step explanation:

We are given the function:

\displaystyle f(x) = \left\{        \begin{array}{ll}            2\cos(\pi x) \text{ for }  x \leq -1 \\ \\          \displaystyle   \frac{2}{\cos(\pi x)}\text{ for } x > -1        \end{array}    \right.

And we want to find:

\displaystyle \lim_{x\to -1}f(x)

So, we need to determine whether or not the limit exists. In other words, we will find the two one-sided limits.

Left-Hand Limit:

\displaystyle \lim_{x\to-1^-}f(x)

Since we are approaching from the left, we will use the first equation:

\displaystyle =\lim_{x\to -1^-}2\cos(\pi x)

By direct substitution:

=2\cos(\pi (-1))=2\cos(-\pi)=2(-1)=-2

Right-Hand Limit:

\displaystyle \lim_{x\to -1^+}f(x)

Since we are approaching from the right, we will use the second equation:

=\displaystyle \lim_{x\to -1^+}\frac{2}{\cos(\pi x)}

Direct substitution:

\displaystyle =\frac{2}{\cos(\pi (-1))}=\frac{2}{\cos(-\pi)}=\frac{2}{(-1)}=-2

So, we can see that:

\displaystyle \displaystyle \lim_{x\to-1^-}f(x)=\displaystyle \lim_{x\to -1^+}f(x) =-2

Since both the left- and right-hand limits exist and equal the same thing, we can conclude that:

\displaystyle \lim_{x \to -1}f(x)=-2

Our answer is A.

8 0
3 years ago
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