All categories

2 years ago

All my question are about math and this test determines my high school classes help if you can please

Mathematics

1 answer:

Mashcka [7]2 years ago

3 0

Answer:

y = x+2

Step-by-step explanation:

When x = 0, y =2

So, y = 0 + m

2 = 0 + m

2 = m

When x = 1, y =3

y = x + m

3 = x + 2

-2 -2

1 = x

You might be interested in

Tara used 8 centimeters of tape to wrap 4 presents. How many presents did Tara wrap if she used 10 centimeters of tape?

Akimi4 [234]

Answer:

She wrapped 5 presents

Step-by-step explanation:

She used 2cm to wrap each present

8cm = 4x

x is the amount of presents

divide by 4

x= 2 cm

Given 10 cm, divide by 2 to determine amount of x, or presents

7 0

3 years ago

The sequence below represents Marisa’s fine at the library for each day that she has an overdue book:

Georgia [21]

The correct answer for the question that is being presented above is this one: "C. f(n) = 0.15n + 0.35." The equation that represents Marisa’s library fine as a function of a book that is n days overdue is f(n) = 0.15n + 0.35

Here are the following choices:
A. f(n) = 0.15n
B. f(n) = 0.50n
C. f(n) = 0.15n + 0.35
D. f(n) = 0.50n + 0.15

4 0

3 years ago

Read 2 more answers

How do I find the integral ∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?

defon

$\int\frac{10}{(x-1)(x^2+9)}\ dx=(*)\\\\\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x-1)}{(x-1)(x^2+9)}\\\\=\frac{Ax^2+9A+Bx^2-Bx+Cx-C}{(x-1)(x^2+9)}=\frac{(A+B)x^2+(-B+C)x+(9A-C)}{(x-1)(x^2+9)}\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1$

$(*)=\int\left(\frac{1}{x-1}+\frac{-x-1}{x^2+9}\right)\ dx=\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)\ dx\\\\=\int\frac{1}{x-1}\ dx-\int\frac{x+1}{x^2+9}\ dx=\int\frac{1}{x-1}-\int\frac{x}{x^2+9}\ dx-\int\frac{1}{x^2+9}\ dx=(**)\\\\\#1\ \int\frac{1}{x-1}\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int\frac{1}{t}\ dt=lnt+C_1=ln(x-1)+C_1$

$\#2\ \int\frac{x}{x^2+9}\ dx\Rightarrow \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=\frac{1}{2}\ du\end{array}\right|\Rightarrow\int\left(\frac{1}{2}\cdot\frac{1}{u}\right)\ du=\frac{1}{2}\int\frac{1}{u}\ du\\\\\\=\frac{1}{2}ln(u)+C_2=\frac{1}{2}ln(x^2+9)+C_2$

$\#3\ \int\frac{1}{x^2+9}\ dx=\int\frac{1}{x^2+3^2}\ dx=\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-\frac{1}{2}ln(x^2+9)+C_2-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3$

$\boxed{=ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C}$

4 0

3 years ago

Please answer my question in the comments

lidiya [134]

Answer: c

Step-by-step explanation:

I would say c because it explains what it's saying and looks like it will show the answer by saying how many more

3 0

3 years ago

Factor 13y 2 - 5yx - 8x 2

sveticcg [70]

Factor the expression and your answer would be
- (5xy + 16x - 26y)
hope this helps