All categories

2 years ago

Y=-3x + 1

Mathematics

1 answer:

Gnesinka [82]2 years ago

8 0

Answer:

rise = -3; run = 1

rise = 3; run = -1

Step-by-step explanation:

The equation is

y = -3x + 1

The equation is in the slope-intercept form,

y = mx + b

m = -3

The slope is -3.

slope = rise/run = -3

We need to write -3 as a fraction to see the rise and the run.

-3 = -3/1 = 3/-1

rise/run = -3/1

rise/run = 3/-1

You can have a rise of -3 and a run of 1, or you can have a rise of 3 and a run of -1. They are both correct.

You might be interested in

What is 5×10 to the power of -3 in as ordinary number

Mumz [18]

Since the exponent of the scientific notation is negative (-3), move the decimal point 3 places to the left. 0.005.

5 0

3 years ago

Will give points 1.. Which of the towers did you select? Draw a sketch of your tower. The height you are given is the vertical d

kati45 [8]

what? for me sheesh ✌️

7 0

3 years ago

Please help!!! .The HCF of 90 and another number is 30. If the LCM of the number is 360, find the other number

Colt1911 [192]

Answer:

The other number is 120.

Step-by-step explanation:

30×360 =90xx

x=(30×360)/90

x=120

(HCFXLCM= Product if the two

numbers)

3 0

1 year ago

17 pointsss!!!!!!! What makes the equation 3(x-6)=8x-3(3x+7 true

almond37 [142]

Answer:

C (-3/4)

Step-by-step explanation:

Decimal form is - 0.75

6 0

3 years ago

Let f be defined by the function f(x) = 1/(x^2+9)

riadik2000 [5.3K]

(a)

$\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}$

Substitute x = 3 tan(t ) and dx = 3 sec²(t ) dt :

$\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt$

$=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}$

(b) The series

$\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}$

converges by comparison to the convergent p-series,

$\displaystyle\sum_{n=3}^\infty\frac1{n^2}$

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}$

converges absolutely, since

$\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}$

That is, ∑ (-1)ⁿ (n ² + 9)/eⁿ converges absolutely because ∑ |(-1)ⁿ (n ² + 9)/eⁿ| = ∑ (n ² + 9)/eⁿ in turn converges by comparison to a geometric series.

5 0

3 years ago