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3 years ago

What are the coordinates for the x and y intercepts of the function 4y - 2x = 24?

2 answers:

8 0

For 4y, I believe the coordinates are (4,6) and the coordinates for 2x are (2,12). This is my best guess. Let me know if I'm wrong.

Alexeev081 [22]3 years ago

6 0

First, let's put this in Slope-intercept form: y = mx + b, where m is the slope and b is the y-intercept.
4y - 2x = 24 Add 2x to both sides
4y = 2x + 24 Divide both sides by 4
y = (1/2)x + 6
Now that we're in slope-intercept we can easily find the y and x intercepts/
b = 6, so the y-intercept equals 6. Plug zero in for x to confirm!

Now we plug 0 in for y to find the x intercept:
0 = (1/2)x + 6 Subtract 6 from both sides
-6 = (1/2) x Multiply both sides by 2
-12 = x
The x intercept is -12

In Conlusion:
x-intercept = -12
y- intercept = 6

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Problem: The mean number of bankruptcies filed per minute in the ` States in a recent year was about two. Find the probability t

tino4ka555 [31]

The number of companies is quite large. That is, n is quite large.
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4 0

3 years ago

Please help me answer this question

avanturin [10]

By direct substitution and simplification, the trigonometric function z = cos (2 · x + 3 · y) represents a solution of the partial differential equation $\frac{\partial^{2} t}{\partial x^{2}} - \frac{\partial^{2} t}{\partial y^{2}} = 5\cdot z$ .

<h3>How to analyze a differential equation</h3>

Differential equations are expressions that involve derivatives. In this question we must prove that a given expression is a solution of a differential equation, that is, substituting the variables and see if the equivalence is conserved.

If we know that $z = \cos (2\cdot x + 3\cdot y)$ and $\frac{\partial^{2} t}{\partial x^{2}} - \frac{\partial^{2} t}{\partial y^{2}} = 5\cdot z$ , then we conclude that:

$\frac{\partial t}{\partial x} = -2\cdot \sin (2\cdot x + 3\cdot y)$

$\frac{\partial^{2} t}{\partial x^{2}} = - 4 \cdot \cos (2\cdot x + 3\cdot y)$

$\frac{\partial t}{\partial y} = - 3 \cdot \sin (2\cdot x + 3\cdot y)$

$\frac{\partial^{2} t}{\partial y^{2}} = - 9 \cdot \cos (2\cdot x + 3\cdot y)$

$- 4\cdot \cos (2\cdot x + 3\cdot y) + 9\cdot \cos (2\cdot x + 3\cdot y) = 5 \cdot \cos (2\cdot x + 3\cdot y) = 5\cdot z$

To learn more on differential equations: brainly.com/question/14620493

#SPJ1

3 0

2 years ago

I need help with these.

sattari [20]

Answer:

See below.

Step-by-step explanation:

5^2 + 12^2 = x^2

25 + 144 = x^2

169 = x^2

x = 13

3^2 + x^2 = [sqrt(10)]^2

9 + x^2 = 10

x^2 = 1

x = 1

1^2 + x^2 = 4^2

1 + x^2 = 16

x^2 = 15

x = sqrt(15)

[sqrt(27)]^2 + x^2 = 6^2

27 + x^2 = 36

x^2 = 9

x = 3

15/5 = c/sqrt(29)

5c = 15 * sqrt(29)

c = 3sqrt(29)

26/13 = x/12

2 = x/12

x = 24

7 0

3 years ago