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3 years ago

The graph of a linear equation passes through three of these points: (0,5), (2, 2), (3, 1), and (4.-1).

Mathematics

1 answer:

sdas [7]3 years ago

7 0

Answer:

(3, 1)

Step-by-step explanation:

plotting these points on a graph, while (3, 1) may look like it's on the line, it is the only point that would involve changing the slope of the line, so that point cannot be on the line.

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Please help!!! I’m going to insert bs equation here *4 + 6 = 10 is an example of an equation.*

Cloud [144]

Answer:

heart exclamation mark

Step-by-step explanation:

you could also maybe make it into a really badly done pokeball

6 0

2 years ago

What is the factorization of 216x12 – 64?

GalinKa [24]

9514 1404 393

Answer:

Step-by-step explanation:

The correct factoring can be found by looking at the first exponent in the second set of parentheses.

The factorization of ...

(a^3 -b^3)

is ...

(a -b)(a^2 +ab +b^2)

Here, you have a=6x^4 and b=4, so the first term in the second parentheses is ...

a^2 = (6x^4)^2 = 6^2·x^(4·2) = 36x^8 . . . . matches the 3rd choice

4 0

2 years ago

2) The equation y = 18x represents the relationship between 2, the number of hours biked, and

Inessa05 [86]

Answer:(2,36)and (3,54)

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Dashna paid a $20 fee for a booth at an art fair so she could sell her ceramic bowls. She will earn $10 for every ceramic bowl s

katovenus [111]

Answer:

The correct answer is D

Step-by-step explanation:

-20 is your y-intercept and your graph will be increasing

6 0

2 years ago

Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

5 0

2 years ago