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3 years ago

What is the equation for slope -2,point (-1,4) in y=Mx+b form

Mathematics

1 answer:

Tema [17]3 years ago

8 0

Answer:

y=-2×+2

Step-by-step explanation:

Hope this helped you out and sorry if it is wrong!

You might be interested in

For the following, write an equivalent expression in standard form: (5x2 + 6x - 2) + 4x (3x – 7).

Roman55 [17]

Answer:

17x^2 -22x -2

Step-by-step explanation:

5x^2 +6x -2 +4x (3x-7)
5x^2 +6x -2 +12x^2 -28x
17x^2 +6x -2 -28x
Answer: 17x^2 -22x -2

Just so you know when I put (^) the number after it is a exponent.

7 0

3 years ago

I NEED to know the answer and steps to figure out, Mark wants to paint a mural. He has 1 1/3 gallons of yellow paint, 1 1/4 gall

emmasim [6.3K]

Well, Mark has

1 1/3 of yellow paint

1 1/4 of green paint and

7/8 of blue paint

now, bear in mind, the yellow is a bucket and 1/3 more, the green one is one bucket and 1/4 more and the blue is 7/8 which is almost 8/8, so is almost a bucketfull.

Then he's going to use 3/4 of each, so, we'd have to subtract 3/4 from each, and there'll be some left still in the paint bucket.

How many will he have left altogether? well, we simply add the leftovers.

$\bf 1\frac{1}{3}\implies \cfrac{1\cdot 3+1}{3}\implies \cfrac{4}{3}
\\\\\\
\cfrac{4}{3}-\stackrel{\downarrow }{\cfrac{3}{4}}\implies \cfrac{16-9}{12}\implies \boxed{\cfrac{7}{12}}\\\\
-------------------------------\\\\
1\frac{1}{4}\implies \cfrac{1\cdot 4+1}{4}\implies \cfrac{5}{4}
\\\\\\
\cfrac{5}{4}-\stackrel{\downarrow }{\cfrac{3}{4}}\implies \cfrac{5-3}{4}\implies \cfrac{2}{4}\implies \boxed{\cfrac{1}{2}}\\\\
-------------------------------\\\\$

$\bf \cfrac{7}{8}-\stackrel{\downarrow }{\cfrac{3}{4}}\implies \cfrac{7-6
}{8}\implies \boxed{\cfrac{1}{8}}\\\\
-------------------------------\\\\
\textit{so the leftovers are }\cfrac{7}{12}+\cfrac{1}{2}+\cfrac{1}{8}\implies \cfrac{14+12+3}{24}\implies \cfrac{29}{24}\implies 1\frac{5}{24}$

6 0

3 years ago

Consider the circle of radius 5 centered at (0, 0). Find an equation of the line tangent to the circle at the point (3, 4) in sl

Wittaler [7]

Answer:

$\displaystyle y= -\frac{3}{4} x + \frac{25}{4}$ .

Step-by-step explanation:

The equation of a circle of radius $5$ centered at $(0,0)$ is:

$x^{2} + y^{2} = 5^{2}$ .

$x^{2} + y^{2} = 25$ .

Differentiate implicitly with respect to $x$ to find the slope of tangents to this circle.

$\displaystyle \frac{d}{dx}[x^{2} + y^{2}] = \frac{d}{dx}[25]$

$\displaystyle \frac{d}{dx}(x^{2}) + \frac{d}{dx}(y^{2}) = 0$ .

Apply the power rule and the chain rule. Treat $y$ as a function of $x$ , $f(x)$ .

$\displaystyle \frac{d}{dx}(x^{2}) + \frac{d}{dx}(f(x))^{2} = 0$ .

$\displaystyle \frac{d}{dx}(2x) + \frac{d}{dx}(2f(x)\cdot f^{\prime}(x)) = 0$ .

That is:

$\displaystyle \frac{d}{dx}(2x) + \frac{d}{dx}\left(2y \cdot \frac{dy}{dx}\right) = 0$ .

Solve this equation for $\displaystyle \frac{dy}{dx}$ :

$\displaystyle \frac{dy}{dx} = -\frac{x}{y}$ .

The slope of the tangent to this circle at point $(3, 4)$ will thus equal

$\displaystyle \frac{dy}{dx} = -\frac{3}{4}$ .

Apply the slope-point of a line in a cartesian plane:

$y - y_0 = m(x - x_0)$ , where

$m$ is the gradient of this line, and
$(x_0, y_0)$ are the coordinates of a point on that line.

For the tangent line in this question:

$\displaystyle m = -\frac{3}{4}$ ,
$(x_0, y_0) = (3, 4)$ .

The equation of this tangent line will thus be:

$\displaystyle y - 4 = -\frac{3}{4} (x - 3)$ .

That simplifies to

$\displaystyle y= -\frac{3}{4} x + \frac{25}{4}$ .

3 0

3 years ago

If f(x) = (4x-3)^2, then f(-2)=? A: -121 B: -22 C: 22 D: 25 E: 121

SOVA2 [1]

The answer is E. 121.

7 0

3 years ago

Can you help me with try these B?

Softa [21]

A. expanded, you get XxXxXxXxXxXxXxX over XxXxXxXxX
b.XxXxX or X³
c. I would just cancel out the 5 Xs on the top line against the 5 Xs on the bottom, leaving X³.

6 0

3 years ago