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3 years ago

Write the number shown on the calculator display in standard form.

Mathematics

1 answer:

frez [133]3 years ago

3 0

Answer:

2 Because visual specialization skills are important to success in so many career

fields, it is helpful to continue to build skill. What is one strategy you can use to

continue to improve your visualization skills?

Please someone help!!!

Step-by-step explanation:

2 Because visual specialization skills are important to success in so many career

fields, it is helpful to continue to build skill. What is one strategy you can use to

continue to improve your visualization skills?

Please someone help!!!

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What is the solution for 5x^2-x+6=0

kati45 [8]

The easiest way to solve this problem is by using the Pythagorean theorem :D

Pythagorean theorem: [-b +- sqrt(b^2 - 4ac)]/2a

5 = a

-1 = b

6 = c

therefore, by plugging these values in!

[-(-1) +- sqrt((-1)^2 - 4(5)(6))]/2(5)

[1 +- sqrt(1 - 120)]/10

Oh so we're using imaginary numbers, so in this case you'll need to know that i = sqrt(-1), so keep this in mind (I can see why this was so difficult)

[1 +- sqrt(119)*i]10

answers:

( 1 + i*sqrt(119) ) / 10

( 1 - i*sqrt(119) ) / 10

7 0

3 years ago

56. How many tangent lines to the curve <img src="https://tex.z-dn.net/?f=y%3Dx%20%2F%28x%2B1%29" id="TexFormula1" title="y=x /(

PIT_PIT [208]

There are 2 tangent lines that pass through the point

$y=\frac{1}{(-1+\sqrt{3)^2} } (x-1)+2$

and

$y=\frac{1}{(-1-\sqrt{3)^2} } (x-1)+2$

Explanation:

Given:

$y=\frac{x}{x+1}$

The point-slope form of the equation of a line tells us that the form of the tangent lines must be:

$y=m(x-1)+2$ $[1]$

For the lines to be tangent to the curve, we must substitute the first derivative of the curve for $m$ :

$\frac{dy}{dx} =\frac{d(x)}{dx}(x+1)-x^\frac{d(x+1)}{dx} \\ \\$

$\frac{dy}{dx} =\frac{x+1-x}{(x+1)^2}$

$\frac{dy}{dx}= \frac{1}{(x+1)^2}$

$m=\frac{1}{(x+1)^2}$ $[2]$

Substitute equation [2] into equation [1]:

$y=\frac{x-1}{(x+1)^2}+2$ $[1.1]$

Because the line must touch the curve, we may substitute $y=\frac{x}{x+1}:$

$\frac{x}{x+1}=\frac{x-1}{(x+1)^2}+2$

Solve for x:

$x(x+1)=(x-1)+2(x+1)^2$

$x^2+x=x-1+2x^2+4x+2$

$x^2+4x+1$

$x\frac{-4±\sqrt{4^2-4(1)(1)} }{2(1)}$

$x=-2$ ± $\sqrt{3}$

$x=-2$ ± $\sqrt{3}$ $and$ $x=-2-\sqrt{3}$

There are 2 tangent lines.

$y=\frac{1}{(-1+\sqrt{3)^2} } (x-1)+2$

and

$y=\frac{1}{(-1-\sqrt{3)^2} } (x-1)+2$

8 0

2 years ago

Given the equation y − 3 = 1/2 (x + 6) in point-slope form, identify the equation of the same line in standard form. )

Doss [256]

Answer:

a: x − 2y = −12

Step-by-step explanation:

the standard form of a line is is usually given as Ax + By = C, so just by deduction you can tell that the correct answer is a.

If you want to do the procedure it would be:

y − 3 = 1/2 (x + 6)

y − 3 = 1/2 x + 3

y - 1/2 x = 3 + 3

y - 1/2 x = 6

(y - 1/2 x = 6) * 2 to get rid of the fraction

2y - x = 12 rearrenge the terms

x - 2y = -12

4 0

3 years ago

Simplify 3(2x - 5). -6x - 15 6x - 15 6x - 8 5x - 2

Alekssandra [29.7K]

You just times it. (if you have a more question pm me). thank you

5 0

3 years ago

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Ordered pair for the point

kupik [55]

Answer:

Step-by-step explanation:

(4, -3)

4 0

3 years ago

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