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RoseWind [281]
3 years ago
8

What is the general form of the equation of the line shown?

Mathematics
1 answer:
Nonamiya [84]3 years ago
4 0

Answer:

x - y = 0

Step-by-step explanation:

The given line is y = x.

We could re-write this as x - y = 0, which is in general form.

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I need help on number 5 this is due at 11.59 thank you :)
zavuch27 [327]
Yes algebra!!! Anyways, the m of y=mx+b is the slope of the graph, this affects the graph by making the line go in a angle either negatively or positivly
B is where the plot lies on the y line (up and down)
3 0
3 years ago
Can someone help me please
goldenfox [79]

Answer:

1. \quad\dfrac{1}{k^{\frac{2}{3}}}\\\\2. \quad\sqrt[7]{x^5}\\\\3. \quad\dfrac{1}{\sqrt[5]{y^2}}

Step-by-step explanation:

The applicable rule is ...

  x^{\frac{m}{n}}=\sqrt[n]{x^m}

It works both ways, going from radicals to frational exponents and vice versa.

The particular power or root involved can be in either the numerator or the denominator. The transformation applies to the portion of the expression that is the power or root.

7 0
3 years ago
The product of (1/2+3/4)(1/2-3/4)(i) and divided by (2-3i) is (multiplying complex numbers)
insens350 [35]
B) because i think its that answer
6 0
3 years ago
I feel like it’s either answer 2 or 4. i just don’t understand the -9 part
harina [27]

Answer:

n < 27

Step-by-step explanation:

let the number be n

\frac{1}{3} n - 3 < 6 \\ add \: 3 \: to \: both \: sides \\  \frac{1}{3}n - 3 + 3  <  6 + 3 \\  \frac{1}{3} n  <  \\  9 \\ multiply \: both \: sides \: by \: 3 \\  \frac{1}{3} n  \times 3< 9 \times 3 \\ n < 27

8 0
3 years ago
A sequence is defined recursively by the following rules: f(1)=3f(n+1)=2⋅f(n)−1 Which of the following statements is true about
Radda [10]

Answer:

f(2)=5

f(5)=33

Step-by-step explanation:

The given formula, that recursively defines the sequence is

f(1) = 3 \\ f(n + 1) = 2f(n) - 1

When n=1, we obtain;

f(1+ 1) = 2f(1) - 1 \\ f(2) = 2 \times 3 - 1 \\ f(2) = 6 - 1 \\ f(2) = 5

When n=2, we get:

f(2+ 1) = 2f(2) - 1 \\ f(3) = 2 \times 5 - 1 \\ f(3) = 10 - 1 \\ f(3) = 9

When n=3,

f(3 + 1) = 2f(3) - 1 \\ f(4) = 2f(3) - 1 \\ f(4) = 2 \times 9 - 1 \\ f(4) = 18 - 1 \\ f(4) = 17

When n=4

f(4 + 1) = 2 f(4) - 1 \\ f(5) = 2 \times 17 - 1 \\ f(5) = 34 - 1 \\ f(5) = 33

When n=5,

f(6) = 65

4 0
2 years ago
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