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2 years ago

Linear function need help ASAP

Mathematics

2 answers:

dolphi86 [110]2 years ago

5 0

Y=1/3x+2 I hope it help you.

ElenaW [278]2 years ago

4 0

Answer:

y=1/3x+2

Step-by-step explanation:

Because it is linear equation in 2 variables

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Find the value of the following expression:

Illusion [34]

Ok, so first you solve whatever is in the parentheses.
(28 • 5-5 • 190)-2•228
28• 5 = 140
Then 5 •190 = 950
Then multiply 950 • 140 = 133,000
Then it should look like this:
133,000-2•228
Multiply 2 • 228 = 456
Then subtract 456 from 133,000.
The answer should come out to 132,544.

3 0

2 years ago

Find the length of ST. St= 5x+2 and TV = 3x+12 <br> Thank you for anyone who does answer(:

valentina_108 [34]

I think you forgot to include that T is the midpoint of SV, but what this would mean is that ST = TV.
5x + 2 = 3x + 12

Subtract 2 from both sides.
5x = 3x + 10

Subtract 3x from both sides.
2x = 10

Divide both sides by 2.
x = 5

Now plug that x into ST.
ST = 5(5) + 2
ST = 25 + 2
ST = 27

4 0

3 years ago

Write 2 ratios that are equivalent to 5/6

JulsSmile [24]

Answer: So, 12 : 30 and 2 : 5 are the two equivalent-ratios of 6 : 15.

Step-by-step explanation:

4 0

2 years ago

Question is attached to image<br> question 9a<br><br> pls help

Ilia_Sergeevich [38]

Answer:

(x-2y){5-(x-2y)}

=(x-2y)(5-x+2y)

8 0

1 year ago

Use a power series to approximate the value of the integral with an error of less than 0.0001. (Round your answer to four decima

otez555 [7]

Answer:

The answer is "0.9461"

Step-by-step explanation:

Given:

$\int^1_0 \frac{sin(x)}{x} dx\\\\$

$\int^1_0 \frac{sin(x)}{x} dx\\\\$ ≈ $\sum^2_{n=0}\frac{(-1)^n x^{2n+1} (-1)^n (x-1)^n}{(2n + 1)!}$

$\therefore$

$\to \sin x = \sum^{\infty}_{n=0} (-1)^n\frac{x^{(2n+1)}}{(2n + 1)!}$

$\because \\\\ \to \frac{\sin x}{x} =\sum^{\infty}_{n=0} (-1)^n \frac{x^{2n+1-1}}{(2n+1)!}\\$

$=\sum^{\infty}_{n=0} (-1)^n \frac{x^{2n}}{(2n+1)!}$

The value is in between 0 and 1 then:

$\to \int^1_0 \frac{sin(x)}{x} = =\sum^{2}_{n=0} \frac{(-1)^n}{ (2n+1) (2n+1)!}$

The above-given series is an alternative series, and it will give an error, when the nth term is bounded by its absolute value, that can be described as follows:

$\to \frac{1}{(2n+3) (2n+3)!}< 0.0001\\\\\to (2n+3) (2n+3)!> 0.0001\\\\\to n\geq 2 \\$

So,

$\int^1_0 \frac{sin(x)}{x} dx\\\\$ ≈ $1 - \frac{1}{3 \cdot 3!}+\frac{1}{5 \cdot 5!}\\\\$

$\approx 1 - \frac{1}{3 \cdot 6}+\frac{1}{5 \cdot 120}\\\\\approx 1 - \frac{1}{18}+\frac{1}{600}\\\\\approx 1 - \frac{1}{18}+\frac{1}{600}\\\\\approx \frac{18-1}{18}+\frac{1}{600}\\\\\approx \frac{17}{18}+\frac{1}{600}\\\\\approx \frac{1700+3}{1800}\\\\ \approx \frac{1703}{1800}\\\\\approx 0.9461$

5 0

2 years ago