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2 years ago

A line includes the points (-7, -3) and (10, -6). What is it’s equation in slope-intercept form?

Mathematics

1 answer:

Vaselesa [24]2 years ago

3 0

7,3 and 10,6 put them together

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What is the value of n in the equation 0=18+128n^2?

lbvjy [14]

Answer:

$\large\boxed{\text{no soltion in the set of real numbers}}\\\boxed{n=-\dfrac{3}{8}i\ \vee\ n=\dfrac{3}{8}i\ \text{in the set of complex numbers}}$

Step-by-step explanation:

$18+128n^2=0\qquad\text{subtract 18 from both sides}\\\\128n^2=-18\qquad\text{divide both sides by 128}\\\\n^2=-\dfrac{18}{128}\\\\n^2=-\dfrac{18:2}{128:2}\\\\n^2=-\dfrac{9}{64}$

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3 years ago

The perimeter of the rectangle below is 184 m .What is the value of y

stiks02 [169]

Step-by-step explanation:

2x + 2y = 184 ---- eqn 1

y = 184 - 2x / 2

y = 92 - x ---- eqn 2

7 0

2 years ago

An arc of a circle is 3πm long, and it subtends an angle of 72° at the center of the circle. Find the radius of the circle.

Artist 52 [7]

Answer:

7.5 m

Step-by-step explanation:

circumference = 360/72 × 3π = 15π

so the radius of the circle =

15π / 2π = 7.5 m

7 0

2 years ago

Find the tangent line equation of the curve at the given point. Y=arcsin(7x) at the point where x=sqrt2/14

Mumz [18]

Answer:

$Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})$

Step-by-step explanation:

The equation of the curve is

$Y = sin^{-1}(7x)$

To find the equation of tangent we need to differentiate this equation w.r.t x

So, differentiating we get

$Y'=\frac{7}{\sqrt{1-49x^2} }$

This would give the slope of the tangent line at any given point of which x coordinate is known. In the present case it is $x = \sqrt{\frac{1}{7} }$

Then slope would accordingly be

$Y'=\frac{7}{\sqrt{1-49/49} }$

= ∞

For, $x = \sqrt{\frac{1}{7} }$ , $Y = sin^{-1}(7/7)= \pi/2$

Equation of tangent line, in the point slope form, would be $Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})$

4 0

3 years ago

Please help I don’t get it

Dmitry [639]

Answer:

ur mom isugly as stiuff

Step-by-step explanation:

vut like u

6 0

3 years ago