Basically, because when you want to find the difference between 2.4 and 1.7 using models, you take away the same amount of squares as if you were finding the difference between 240 and 170 using models.

3 0

3 years ago

If cosA=4/5 and cosB=3/5 then find a,b,c

aliya0001 [1]

Answer:

a = 3, b = 4, c = 5

Step-by-step explanation:

Assuming we're working with a right triangle, where c is the hypotenuse, then using the definition of the cosine being the adjacent side over the hypotenuse, then we know:

a = 3, because it is the side adjacent to B

b = 4, because it is the side adjacent to A

c = 5, because it is the denominator in bot fractions

This of course assumes that there is no additional ratio in place. For example, if the lengths were instead 8, 6 and 10 respectively, then the cosines given would still be 4/5 and 3/5. Truthfully these only tell relative sizes of the sides, and not their absolute sizes.

3 0

2 years ago

Read 2 more answers

bobby delivers newspapers in the neighborhood. in addition to a weekly salary, he earns tips from the people he delivers to . if

iVinArrow [24]

17% of $200 is $34

Hope that helps.

6 0

3 years ago

Finding an Equation of a tangent Line in Exercise, find an equation of the tangent line to the graph of the function at the give

frutty [35]

Answer:

$y=\dfrac{3x}{e}+\dfrac{4}{e}$

this is the equation of the tangent at point (-1,1/e)

Step-by-step explanation:

to find the tangent line we need to find the derivative of the function g(x).

$g(x) =e^{x^3}$

we know that $\frac{d}{dx}(e^{f(x)})=e^{f(x)}f'(x)$

$g'(x) =e^{x^{3}}(3 x^{2})$

$g'(x) =3 x^{2} e^{x^{3}}$

this the equation of the slope of the curve at any point x and it also the slope of the tangent at any point x. hence, g'(x) can be denoted as 'm'

to find the slope at (-1,1/e) we'll use the x-coordinate of the point i.e. x = -1

$m =3 (-1)^{2} e^{(-1)^{3}}\\m =3e^{-1}\\m=\dfrac{3}{e}$

using the equation of line:

$(y-y_1)=m(x-x_1)$

we'll find the equation of the tangent line.

here (x1,y1) =(-1,1/e), and m = 3/e

$(y-\dfrac{1}{e})=\dfrac{3}{e}(x+1)\\y=\dfrac{3x}{e}+\dfrac{3}{e}+\dfrac{1}{e}\\$

$y=\dfrac{3x}{e}+\dfrac{4}{e}$

this is the equation of the tangent at point (-1,1/e)

3 0

3 years ago