All categories

3 years ago

What percent is 25 is 12

Mathematics

2 answers:

Ray Of Light [21]3 years ago

8 0

Answer: 48%

Step-by-step explanation:

I believe you meant, what percent of 25 is 12.

In that case, you would do 12/25 which equals 0.48

To convert a decimal into a percent, you would move the decimal point over 2 digits to the right. That would make 0.48 into 48%

So 48% is the answer.

Hope this helps :)

Lana71 [14]3 years ago

5 0

Answer:

208.3333333333%

Step-by-step explanation:

25 /12 x 100% = 208.3333333333%

mark me brainliest please

You might be interested in

Algebra 1 pls help ty

olchik [2.2K]

Answer is D. Good luck

3 0

3 years ago

Read 2 more answers

Explain the difference in meaning between 1-3 and -3

sveticcg [70]

Answer:

The difference in meaning is that

-3 is a signed value. It lies to the left of zero

+3 is also a signed value, and since it is positive, it lies to the right of zero

|-3| is just the physical distance (left or right) of -3 from zero. Since we live in a physical world, all distances are positive, so its signed value is 3.

Since +3 is also 3 units from zero, |+3| = 3

|-3| = |+3| = 3 since both are three units away from zero.

So, when using absolute values as numbers, rather than distances,

|x| = x if x is not negative

|x| = -x if x is negative

7 0

3 years ago

Nine increased by four times a number

Stells [14]

Answer:

9+4x

Step-by-step explanation:

Is this what you want?

8 0

4 years ago

If a=1 and b=-3, what is the value of |a-2| + |2-b|?

inna [77]

It would be 6. If you substitute 1 and -3 into their assigned letters, you would get 6 as your final answer. the two lines mean it is an absolute value, which means any negative number would turn into a positive number

8 0

3 years ago

Read 2 more answers

Find an equation of the tangent line to the hyperbola x2 a2 − y2 b2 = 1 at the point (x0, y0).

nalin [4]

The equation of the tangent line is $y = \frac{b^{2} x_{0}x }{a^{2} y_{0} } - \frac{b^{2} x_{0}^{2} }{a^{2} y_{0} } + y_{0}$

To find the tangent to the hyperbola $\frac{x^{2} }{a^{2} } - \frac{y^{2} }{b^{2} }$ at (x₀, y₀), we differentiate the equation implicitly to find the equation of the tangent at (x₀, y₀).

So, $\frac{d}{dx} (\frac{x^{2} }{a^{2} } - \frac{y^{2} }{b^{2} }) = \frac{d0}{dx}\\\frac{d}{dx} \frac{x^{2} }{a^{2} } - \frac{d}{dx}\frac{y^{2} }{b^{2} }= 0\\\frac{2x }{a^{2} } - \frac{dy}{dx}\frac{2y }{b^{2} } = 0\\\frac{2x }{a^{2} } = \frac{dy}{dx}\frac{2y }{b^{2} } \\\frac{dy}{dx} = \frac{b^{2}x }{a^{2}y }$

So, at (x₀, y₀)

$\frac{dy}{dx} = \frac{b^{2} x_{0} }{a^{2} y_{0} }$

So, the equation of the tangent line is gotten from the standard equation of a line in point-slope form

So, $\frac{y - y_{0} }{x - x_{0} } = \frac{b^{2} x_{0} }{a^{2} y_{0} } \\y - y_{0} = \frac{b^{2} x_{0} }{a^{2} y_{0} }(x - x_{0}) \\y = \frac{b^{2} x_{0} }{a^{2} y_{0} }(x - x_{0}) + y_{0} \\y = \frac{b^{2} x_{0}x }{a^{2} y_{0} } - \frac{b^{2} x_{0}^{2} }{a^{2} y_{0} } + y_{0}$

So, the equation of the tangent line is $y = \frac{b^{2} x_{0}x }{a^{2} y_{0} } - \frac{b^{2} x_{0}^{2} }{a^{2} y_{0} } + y_{0}$

Learn more about equation of tangent line here:

brainly.com/question/12561220

3 0

3 years ago