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JulijaS [17]
3 years ago
15

Write the equation of a line with slope 3 and y-intercept −3. PLS I NEED ASAP

Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
7 0
The equation of the line is Y = 3x + (-3).
You might be interested in
What are the domain, range, and asymptote of h(x) = (0.5)x – 9?
Rzqust [24]

Let's solve for h

hx = 0.5x - 9

Step 1: Divide both sides by x.

hx/x = 0.5x - 9/ x

Answer = h = 0.5x - 9/ x

8 0
2 years ago
I’m need this worked out step by step by tonight
Pie

9514 1404 393

Answer:

  -3 ≤ x ≤ 19/3

Step-by-step explanation:

This inequality can be resolved to a compound inequality:

  -7 ≤ (3x -5)/2 ≤ 7

Multiply all parts by 2.

  -14 ≤ 3x -5 ≤ 14

Add 5 to all parts.

  -9 ≤ 3x ≤ 19

Divide all parts by 3.

  -3 ≤ x ≤ 19/3

_____

<em>Additional comment</em>

If you subtract 7 from both sides of the given inequality, it becomes ...

  |(3x -5)/2| -7 ≤ 0

Then you're looking for the values of x that bound the region where the graph is below the x-axis. Those are shown in the attachment. For graphing purposes, I find this comparison to zero works well.

__

For an algebraic solution, I like the compound inequality method shown above. That only works well when the inequality is of the form ...

  |f(x)| < (some number) . . . . or ≤

If the inequality symbol points away from the absolute value expression, or if the (some number) expression involves the variable, then it is probably better to write the inequality in two parts with appropriate domain specifications:

  |f(x)| > g(x)   ⇒   f(x) > g(x) for f(x) > 0; or -f(x) > g(x) for f(x) < 0

Any solutions to these inequalities must respect their domains.

8 0
2 years ago
Write a quadratic function whose graph has a vertex of (3,2) and passes through the point (4,7). f(x)=
agasfer [191]

Answer:

y=5(x-3)^2+2

Step-by-step explanation:

You plug the point into f(x)=a(x-3)^2+2... that already has already been solved for the vertex that you want. Then you swap it out for the solution you have solved for.

7 0
2 years ago
PLZZZZZZZ NEED HELP!!!!!!
Maurinko [17]
I think it is $30 since it has the highest probability.
7 0
3 years ago
What is the domain and derivative of f(x)=ln(2xsqrt(2+x))?
harkovskaia [24]
Domain is the numbers yo can use
we has a sqrt and a ln
we cant have any neagtive ln's or sqrts
x cannot be negative
it cannot be 0 either
domain is all numbers bigger than 0


deritivive
apply chain rule
dy/dx(f(g(x))=f'(g(x))g'(x)
and also
dy/dx (lnx)=1/x
and
dy/dx(f(x)g(x))=f'(x)g(x)+g'(x)f(x)
and from experience
dy/dx(sqrtx)=1/(2√x)


so

dy/dx(ln(2xsqrt(2+x)))=
(1/(2xsqrt(2+x))(2)(sqrt(2+x)+(x/(2sqrt(2+x))=
(3x+4)/(2x(x+2))



domain is all real numbers greater than 0
derivitive is \frac{3x+4}{2x(x+2)}
7 0
3 years ago
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