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xxTIMURxx [149]
3 years ago
15

Find the shortest distance between the point (-6,4) and the line y = -2x + 7

Mathematics
1 answer:
Black_prince [1.1K]3 years ago
3 0
Use the formula.much easier.

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Multiply 3x (2x - 1)​
Volgvan

Answer:

6x^2 - 3x

6x^2-3x

Step-by-step explanation:

3x (2x-1)

multiply 3x by 2x -> 6x^2

multiply 3x by -1 -> -3x

6 0
3 years ago
Read 2 more answers
How would I solve this?
trasher [3.6K]

You had the right idea using the Pythagorean theorem to solve for b.

Problem is for that triangle to work, the 5 and the 2√2 would have to switch places. The length of a leg cannot be larger than the length of the hypotenuse for it to truly be a right triangle.

Pythagorean theorem only works for the right triangles. Only way to "solve this problem would be to bring in complex numbers.

5² + b² = (2√2)²

25 + b² = 2²(√2)²

25 + b² = 4(2)

25 + b² = 8

b² = 8 - 25

b² = - 17

b = √-17

b= (√17i)

Then the problem with THIS is a measurement/distance cannot be negative... which goes against exactly what that complex number i is.

8 0
3 years ago
If f(x)=x^3-x+2, then (f^-1)'(2)
yawa3891 [41]

Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,

f\left(f^{-1}(x)\right) = x

\implies f^{-1}(x)^3 - f^{-1}(x) + 2 = x

which is a cubic polynomial in f^{-1}(x) with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).

Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.

f'(x) = 3x² - 1 = 0   ⇒   x = ±1/√3

So, we have three subsets over which f(x) can be considered invertible.

• (-∞, -1/√3)

• (-1/√3, 1/√3)

• (1/√3, ∞)

By the inverse function theorem,

\left(f^{-1}\right)'(b) = \dfrac1{f'(a)}

where f(a) = b.

Solve f(x) = 2 for x :

x³ - x + 2 = 2

x³ - x = 0

x (x² - 1) = 0

x (x - 1) (x + 1) = 0

x = 0   or   x = 1   or   x = -1

Then \left(f^{-1}\right)'(2) can be one of

• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);

• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or

• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)

6 0
2 years ago
What is the solution to the linear equation ?
lawyer [7]

Answer:

A

<em>brainliest is much appreciated!</em>

Step-by-step explanation:

hope this helps!

5 0
3 years ago
Read 2 more answers
One computer has a mass 0.8 and another has a mass of 800 grams compare the masses of the computer.use &gt; &lt; = to make a tru
Artyom0805 [142]
Since 800 is greater than 0.8 your statement would be 800>0.8
4 0
3 years ago
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