What is the number written in standard form :six million , seven hundred thousand,twenty

What is the slope intercept from for 16x+9y=40

Doss [256]

Answer:

$y = (- \frac{16}{9})x + \frac{40}{9}$ is the slope intercept form for 16x+9y=40

Step-by-step explanation:

Here, the given equation is 16 x + 9 y = 40

Now, the slope intercept form of an equation is y = mx + c,

here m = slope of the equation, b = y - intercept

So, now 16x + 9y = 40 on simplification gives

9y = 40 - 16x

or $y = \frac{40 - 16x}{9} = \frac{40}{9} - \frac{16x}{9}$

or, $y = (- \frac{16}{9})x + \frac{40}{9}$ is the slope intercept form of the given equation.

Here, m = (-16/9) and y- intercept = 40/9

7 0

3 years ago

HELP PLEASEEEEEEEEEEEEEE

vazorg [7]

Answer:

-2/4, 8/40, 21/30, 9, 9 20/25

Step-by-step explanation:

9/20/25 is a bit greater than 9.

21/30 is 9 away from a whole 8/40 is 32 away from 40 -2/4 is way less because its on the negative side which is like so far away from the other numbers.

7 0

3 years ago

Prove that :( 1 + 1/<img src="https://tex.z-dn.net/?f=tan%5E%7B2%7DA" id="TexFormula1" title="tan^{2}A" alt="tan^{2}A" align="ab

Aliun [14]

Answer:

See explanation

Step-by-step explanation:

Simplify left and right parts separately.

$\left(1+\dfrac{1}{\tan^2A}\right)\left(1+\dfrac{1}{\cot ^2A}\right)\\ \\=\left(1+\dfrac{1}{\frac{\sin^2A}{\cos^2A}}\right)\left(1+\dfrac{1}{\frac{\cos^2A}{\sin^2A}}\right)\\ \\=\left(1+\dfrac{\cos^2A}{\sin^2A}\right)\left(1+\dfrac{\sin^2A}{\cos^2A}\right)\\ \\=\dfrac{\sin^2A+\cos^2A}{\sin^2A}\cdot \dfrac{\cos^2A+\sin^A}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A}\cdot \dfrac{1}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A\cos^2A}$