Well, I can't explain how to solve but the answer is (2,8)
To determine which system of equations would have the same solution, we evaluate each system of equations.
System 1 4x − 5y = 2, 3x − y = 8
x = 38/11
y = 26/11
<span>System 2 4x − 5y = 2, 3x − 2y = 1
x = 1/7
y = -2/7
System 3 4x − 5y = 2, 3x − 8y = 4
x = -4/17
y = -10/17
System 4 4x − 5y = 2, 10x − 9y = 4
x = 1/7
y = -2/7
</span><span>
Therefore, the correct answer is option 3. </span><span>System 2 and system 4 are equal, because the second equation in system 4 is obtained by adding the first equation in system 2 to two times the second equation in system 2.
4x− 5y = 2 2( 3x − 2y = 1)
----------------------- 10x - 9y = 4</span>
N = 273
273 / 7 = 39
Dividend / divisor = quotient
Work Backwards:
n/7 = 39
n = 39 X 7
n = 273
Answer:
f(x) does not have an inverse function, because its graph fails the horizontal line test. ⇒ A
Step-by-step explanation:
The vertical line test is used to show that the graph represents a
function or not
- If the vertical line crosses the graph in more than one point, then the graph does not represent a function
- If the vertical line crosses the graph in only one point in different positions, then the graph represents a function
The horizontal line test is used to show that the function has an
inverse or not
- If the horizontal line crosses the graph of a function in more than one point, then the function has no inverse
- If the horizontal line crosses the graph of a function in only one point in different positions, then the function has an inverse
Let us look at the given graph
→ The graph represents a parabola, which represents the function f(x)
∵ Any horizontal line drawn will cross the graph in more than one point
∴ f(x) has no inverse
∴ f(x) does not have an inverse function, because its graph fails the
horizontal line test.
well, we'll first off put the point AC in component form by simply doing a subtraction of C - A, multiply that by the fraction 2/3, and that result will get added to point A, to get point B.
![\bf \textit{internal division of a segment using a fraction}\\\\ A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{5})\qquad C(\stackrel{x_2}{4}~,~\stackrel{y_2}{-4})~\hfill \frac{2}{3}\textit{ of the way from }A\to C \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_2}{4}-\stackrel{x_1}{(-2)}, \stackrel{y_2}{-4}-\stackrel{y_1}{5})\implies (4+2,-9) \stackrel{\textit{component form of segment AC}}{\qquad \implies \qquad (6,-9)} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Binternal%20division%20of%20a%20segment%20using%20a%20fraction%7D%5C%5C%5C%5C%20A%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B5%7D%29%5Cqquad%20C%28%5Cstackrel%7Bx_2%7D%7B4%7D~%2C~%5Cstackrel%7By_2%7D%7B-4%7D%29~%5Chfill%20%5Cfrac%7B2%7D%7B3%7D%5Ctextit%7B%20of%20the%20way%20from%20%7DA%5Cto%20C%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_2%7D%7B4%7D-%5Cstackrel%7Bx_1%7D%7B%28-2%29%7D%2C%20%5Cstackrel%7By_2%7D%7B-4%7D-%5Cstackrel%7By_1%7D%7B5%7D%29%5Cimplies%20%284%2B2%2C-9%29%20%5Cstackrel%7B%5Ctextit%7Bcomponent%20form%20of%20segment%20AC%7D%7D%7B%5Cqquad%20%5Cimplies%20%5Cqquad%20%286%2C-9%29%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
