All categories

3 years ago

Use the slope-intercept method to graph the equation y=2x-4. Is (3,2) a solution?

Mathematics

2 answers:

Andru [333]3 years ago

8 0

Yes, (3, 2) is a solution to y = 2x - 4. (:

jeka57 [31]3 years ago

3 0

The ordered pair (3,2) is a solution.
To graph the function: first plot the y-intercept at -4. Next from the y-intercept, count up 2 units and then right one unit and plot a point, continue to do this until you have several points plotted (at least 4). Next using a straight-edge, draw a line through the points.

You might be interested in

ASAP! BRAINLIEST for RIGHT answer!!!

AnnZ [28]

I'm probs wrong but either A or D

3 0

4 years ago

Read 2 more answers

Pllssss help what is 717 divided by 9 and 85 divided 8 and 200 divided by 5 please and I need an answer today

sasho [114]

Answer:

717 divided by 9=239/3

85 divided by 8=85/8

200 divided by 5=40

Step-by-step explanation:

7 0

3 years ago

A cabinet set was sold at RM 9500 after a markup of 6.5%. Find the original price and the markup amount.

LiRa [457]

Answer:

1) The original price is 8920 2) The mark up is 580

Step-by-step explanation:

!)We need to find out the initial price, suppose it is equal to 100 percents. The price was raised and it has reached 100+6.5= 106.5 percents . 106.5 percents from initial price is the new price (RM 9500). Find out the value of one percent

9500:106.5= 89.2 (approximately).

2)Then find the original price multiplying the value of percent by 100. 89.2*100= 8920 - initial price

3)Then mark up is a remainder between old price and new price . It is 9500-8920= 580.

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%20%5Cdisplaystyle%20%5Cint%20%5Climits_%7B0%7D%5E%7B%20%5Cfrac%7B%20%5Cpi%7D%7B2%7D%20%7D%

murzikaleks [220]

Let $x = \arcsin(y)$ , so that

$\sin(x) = y$

$\tan(x)=\dfrac y{\sqrt{1-y^2}}$

$dx = \dfrac{dy}{\sqrt{1-y^2}}$

Then the integral transforms to

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_{y=\sin(0)}^{y=\sin\left(\frac\pi2\right)} \frac{y}{\sqrt{1-y^2}} \ln(y) \frac{dy}{\sqrt{1-y^2}}$

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy$

Integrate by parts, taking

$u = \ln(y) \implies du = \dfrac{dy}y$

$dv = \dfrac{y}{1-y^2} \, dy \implies v = -\dfrac12 \ln|1-y^2|$

For 0 < y < 1, we have |1 - y²| = 1 - y², so

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = uv \bigg|_{y\to0^+}^{y\to1^-} + \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy$

It's easy to show that uv approaches 0 as y approaches either 0 or 1, so we just have

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy$

Recall the Taylor series for ln(1 + y),

$\displaystyle \ln(1+y) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n y^n$

Replacing y with -y² gives the Taylor series

$\displaystyle \ln(1-y^2) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n (-y^2)^n = - \sum_{n=1}^\infty \frac1n y^{2n}$

and replacing ln(1 - y²) in the integral with its series representation gives

$\displaystyle -\frac12 \int_0^1 \frac1y \sum_{n=1}^\infty \frac{y^{2n}}n \, dy = -\frac12 \int_0^1 \sum_{n=1}^\infty \frac{y^{2n-1}}n \, dy$

Interchanging the integral and sum (see Fubini's theorem) gives

$\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy$

Compute the integral:

$\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy = -\frac12 \sum_{n=1}^\infty \frac{y^{2n}}{2n^2} \bigg|_0^1 = -\frac14 \sum_{n=1}^\infty \frac1{n^2}$