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3 years ago

12

What is the slope-intercept form of the function described by this table?

1 answer:

Vikentia [17]3 years ago

8 0

Y=5x - 3 Because the difference in the y values are 5, the rise is 5. That is the TOP of the slope. The difference in the x values is 1, that is the run or the BOTTOM of the slope. So m = 5/1. Using y = mx + b 2 = 5(1) + b -3 = b Therefore, y = 5x -3

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Find the remaining trigonometric functions of θ based on the given information.csc θ = 257 and cos θ < 0

katrin [286]

Cos(<span>θ) < 0, so we know it would be in Quadrant 2 or 3
then csc(</span>θ) = 257, but csc(θ) = $\frac{1}{sin(θ)}$ = 257
==> sin(<span>θ) = $\frac{1}{257}$
it is positive, so now we can determine that is in Quadrant 2

sin(</span>θ) = opp./hyp both of opp and hyp are positive but adj suppose to negative because that way it leads the cos(<span>θ) < 0
</span>cos(<span>θ) = adj/hyp
</span>Pythereom to find the adj: $257^{2} - 1^{2} = adj ==\ \textgreater \ adj \sqrt{ 257^{2} - 1^{2} } ==\ \textgreater \ adj = 16 \sqrt{258}$

cosθ = $\frac{16 \sqrt{258} }{257} 
$
tanθ = $\frac{1}{16 \sqrt{258} }$
cos<span>θ = </span> $16 \sqrt{258}$

6 0

3 years ago

In this problem we consider an equation in differential form Mdx+Ndy=0. (4x+2y)dx+(2x+8y)dy=0 Find My= 2 Nx= 2 If the problem is

zheka24 [161]

Answer:

$f(x,y)=2x^2+4y^2+2xy=C_1\\\\Where\\\\y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

Step-by-step explanation:

Let:

$M(x,y)=4x+2y\\\\and\\\\N(x,y)=2x+8y$

This is and exact equation, because:

$\frac{\partial M(x,y)}{\partial y} =2=\frac{\partial N}{\partial x}$

So, define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x} =M(x,y)\\\\and\\\\\frac{\partial f(x,y)}{\partial y} =N(x,y)$

The solution will be given by:

$f(x,y)=C_1$

Where C1 is an arbitrary constant

Integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y):

$f(x,y)=\int\ {4x+2y} \, dx =2x^2+2xy+g(y)$

Where g(y) is an arbitrary function of y.

Differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y} =2x+\frac{d g(y)}{dy}$

Substitute into $\frac{\partial f(x,y)}{\partial y} =N(x,y)$

$2x+\frac{dg(y)}{dy} =2x+8y\\\\Solve\hspace{3}for\hspace{3}\frac{dg(y)}{dy}\\\\\frac{dg(y)}{dy}=8y$

Integrate $\frac{dg(y)}{dy}$ with respect to y:

$g(y)=\int\ {8y} \, dy =4y^2$

Substitute g(y) into f(x,y):

$f(x,y)=2x^2+4y^2+2xy$

The solution is f(x,y)=C1

$f(x,y)=2x^2+4y^2+2xy=C_1$

Solving y using quadratic formula:

$y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

4 0

3 years ago

Need help fast!! Please help

Luden [163]

Answer:

I am pretty sure it is 10.5

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

How to solve this problem?<br> -3 d +7 > 25

sp2606 [1]

Answer to your question:

D = d < -6

8 0

3 years ago

Read 2 more answers

You have $3.20 in quarters and dimes. You have four fewer quarters than dimes. What would the associated system of equations loo

Harlamova29_29 [7]

The equation is
Dimes is x

x+x-4=3.20

5 0

3 years ago