All categories

2 years ago

3) Write in expanded form: 12, 245

Mathematics

2 answers:

spin [16.1K]2 years ago

5 0

Twelve, two hundred and fourty five

Alchen [17]2 years ago

3 0

Answer:12,000+200+40+5

Step-by-step explanation:

You might be interested in

Please help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Divide

IgorC [24]

Answer:

The answer is option 2.

Step-by-step explanation:

Firstly, you must factorize the possible expressions :

$\frac{ {x}^{2} - 2x - 15 }{x - 3} \div \frac{x - 5}{ {x}^{2} - 9 }$

$= \frac{(x - 5)(x + 3)}{x - 3} \div \frac{x - 5}{(x - 3)(x + 3)}$

Next you have to divide by converting to multiplication :

$= \frac{(x - 5)(x + 3)}{x - 3} \times \frac{(x - 3)(x + 3)}{x - 5}$

Lastly, you can cut out the similar expressions :

$= \frac{(x - 5)(x + 3)(x - 3)(x + 3)}{(x - 3)(x - 5)}$

$= {(x + 3)}^{2}$

7 0

2 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:

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

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

$x^2+4x+1$

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

$x=-2$ ± $\sqrt{3}$

$x=-2$ ± $\sqrt{3}$ <em> </em> $and$ $x=-2-\sqrt{3}$

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

A swimmer is racing to the other side of the

kiruha [24]

Either B or C. They look exactly the same but I can’t tell because the photo is blurry

y intercept is 75 and x intercept is 75/2.5=30

4 0

2 years ago

A ball is thrown from an initial height of 2 feet with an initial upward velocity of 31 ft/s. The ball's height h (in feet) afte

Vlad1618 [11]

A quadratic equation is in the form of ax²+bx+c. The time at which the height of the ball is 16 feets is 0.717 seconds and 1.221 seconds.

<h3>What is a quadratic equation?</h3>

A quadratic equation is an equation whose leading coefficient is of second degree also the equation has only one unknown while it has 3 unknown numbers. It is written in the form of ax²+bx+c.

The complete question is:

A ball is thrown from an initial height of 2 feet with an initial upward velocity of 31 ft/s. The ball's height h (in feet) after 7 seconds is given by the following, h=2+31t-16t². Find all values of t for which the ball's height is 16 feet. Round your answer(s) to the nearest hundredth.