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Anestetic [448]
2 years ago
11

SOMEOne PLS EXPLAIN VERTICAL ASYMPTOTES TO ME IM DYING- i need to know how u get the asymptote(explanation) and answer T^TTTTTTT

TT

Mathematics
1 answer:
Kitty [74]2 years ago
4 0
A vertical asymptote is what you get when you try to divide by 0. To find where you get these, you need to look at the denominator and what values of x will make the denominator equal to 0.

In your denominator, you have (x+7)(x-5)(x-3).
What values of x makes (x+7)(x-5)(x-3)=0?
If x = -7, if x = 5, or if x = 3, then that entire expression will equal zero. (Same idea as when you solve equations by factoring.

Now the only place this can get trickier is if one of those factors — one of (x+7), (x-5), or (x-3) — also appears in the numerator. If that happens, then it’s more involved whether you have an asymptote or not. But that doesn’t happen in this example.

So the short version: Asymptotes happen when you try to divide by zero. Dividing by zero is not a good thing. So you just ask yourself, “What will make the denominator 0?”
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What is six times three divided by 2
tia_tia [17]

Answer:

9

Step-by-step explanation:

(6*3)÷2

18÷2

9 answer

8 0
2 years ago
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3/6 times 2/6. Please show step by step directions I NEED HELP ASAP
viva [34]

To solve, simply layout and equation.

3/6*2/6.

Step 1: Simply 3/6.

1/2*2/6

Step 2: Simplify 2/6.

1/2*1/3

Step 3: Multiply.

1/2*1/3

1/6.

So, the answer to this problem is 1/6.

7 0
3 years ago
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Is the triangle obtuse, acute, equilateral or right?
Stells [14]

9514 1404 393

Answer:

  obtuse

Step-by-step explanation:

The law of cosines tells you ...

  b² = a² +c² -2ac·cos(B)

Substituting for a²+c² using the given equation, we have ...

  b² = b²·cos(B)² -2ac·cos(B)

We can subtract b² to get a quadratic in standard form for cos(B).

  b²·cos(B)² -2ac·cos(B) -b² = 0

Solving this using the quadratic formula gives ...

  \cos(B)=\dfrac{-(-2ac)\pm\sqrt{(-2ac)^2-4(b^2)(-b^2)}}{2b^2}\\\\\cos(B)=\dfrac{ac}{b^2}\pm\sqrt{\left(\dfrac{ac}{b^2}\right)^2+1}

The fraction ac/b² is always positive, so the term on the right (the square root) is always greater than 1. The value of cos(B) cannot be greater than 1, so the only viable value for cos(B) is ...

  \cos(B)=\dfrac{ac}{b^2}-\sqrt{\left(\dfrac{ac}{b^2}\right)^2+1}

The value of the radical is necessarily greater than ac/b², so cos(B) is necessarily negative. When cos(B) < 0, B > 90°. The triangle is obtuse.

4 0
2 years ago
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In triangle ABC, D is the midpoint of side AC and E is the midpoint of side BC.
cestrela7 [59]

Answer:

14

Step-by-step explanation:

Since all sides are half of greater triangle.Then 7 × 2 = 14 is length of greater hypotenuse.

Suppose two triangles with similar conditions

1^2+1^2=2

Hypotenuse=2^1/2

We assume another triangle which is exactly double of previous triangle I.e

2^2+2^2=8

Hypotenuse=2×2^1/2

Which is exactly twice of previous triangle.

Hence proved hyootenuse is double of smaller triangle I.e 14

8 0
3 years ago
A^2(64a^2-3)/3=27-4a^2/4
frutty [35]
The answer is a = 3/4 = 0.75

First get rid of the paranthesis,
{a}^{2} (64 {a}^{2}  - 3) = 64 {a}^{4} -  3 {a}^{2}
Then set the denominators equal:
\frac{ 256 {a}^{4}  - 12 {a}^{2}}{12}  =  \frac{81 - 12 {a}^{2} }{12}
Then remove the denominators and solve:
256 {a}^{4}  - 12 {a}^{2}  = 81 - 12 {a}^{2}
Eliminate -12a^2 by adding 12a^2 to both sides:
256 {a}^{4}  = 81
Take the fourth root of them or take the square root twice:
\sqrt[4]{256 {a}^{4} }  =  \sqrt[4]{81}   \\  4a = 3
Divide both sides by 4:
a =  \frac{3}{4}  = 0.75
7 0
2 years ago
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