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3 years ago

What is the equation of the line in the slope intercept form

Mathematics

2 answers:

Grace [21]3 years ago

7 0

Answer:

Equation of line; y = - 6x

Step-by-step explanation:

As we can see from this graph, point ( 0, 0 ) lying on this graph intersects the y - axis such that it forms a y - intercept of 0;

At the same time we can note that the change in y / change in x, in other words the slope, differs by a rise of 6 / run of - 1, 6 / - 1 being a slope of - 6;

If this equation is in slope - intercept form ⇒

y = a * x + b, where a ⇒ slope, and b ⇒ y - intercept,

Equation of line; y = - 6x

sergejj [24]3 years ago

3 0

Answer:

y = -6x

Step-by-step explanation:

The slope intercept form of a line is

y = mx+b where m is the slope and b is the y intercept

the y intercept is 0 since it crosses the y axis at 0

To find the slope we need to points (0,0) and (10,-60)

m = (y2-y1)/(x2-x1)

= (-60-0)/(10-0)

= -60/10

= -6

The equation for the line is

y = -6x +0

y = -6x

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rusak2 [61]

Answer:

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Step-by-step explanation:

Do Pemdas to find order of operations:

(264*675)+ 1239-56+887

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(178200)+2070

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3 years ago

A spacecraft is traveling with a velocity of v0x = 5320 m/s along the +x direction. Two engines are turned on for a time of 739

OlgaM077 [116]

Answer:

The velocities after 739 s of firing of each engine would be 6642.81 m/s in the x direction and 5306.02 in the y direction

Step-by-step explanation:

For a constant acceleration: $v_{f}=v_{0}+at$ , where $v_{f}$ is the final velocity in a direction after the acceleration is applied, $v_{0}$ is the initial velocity in that direction before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied.
Then for the x direction it is known that the initial velocity is $v_{0x} =$ 5320 m/s, the acceleration (the applied by the engine) in x direction is $a_{x}$ 1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the is 739 s. Then: $v_{fx}=v_{0x}+a_{x}t=5320\frac{m}{s} +1.79\frac{m}{s^{2} }*739s=6642.81\frac{m}{s}$
In the same fashion, for the y direction, the initial velocity is $v_{0y} =$ 0 m/s, the acceleration in y direction is $a_{y}$ 7.18 m/s2, and the time is the same that in the x direction, 739 s, then for the final velocity in the y direction: $v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}$

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3 years ago

HELP! Find the value of sin 0 if tan 0 = 4; 180 < 0< 270

BabaBlast [244]

Hi there! Use the following identities below to help with your problem.

$\large \boxed{sin \theta = tan \theta cos \theta} \\ \large \boxed{tan^{2} \theta + 1 = {sec}^{2} \theta}$

What we know is our tangent value. We are going to use the tan²θ+1 = sec²θ to find the value of cosθ. Substitute tanθ = 4 in the second identity.

$\large{ {4}^{2} + 1 = {sec}^{2} \theta } \\ \large{16 + 1 = {sec}^{2} \theta } \\ \large{ {sec}^{2} \theta = 17}$

As we know, sec²θ = 1/cos²θ.

$\large \boxed{sec \theta = \frac{1}{cos \theta} } \\ \large \boxed{ {sec}^{2} \theta = \frac{1}{ {cos}^{2} \theta} }$

And thus,

$\large{ {cos}^{2} \theta = \frac{1}{17}} \\ \large{cos \theta = \frac{ \sqrt{1} }{ \sqrt{17} } } \\ \large{cos \theta = \frac{1}{ \sqrt{17} } \longrightarrow \frac{ \sqrt{17} }{17} }$

Since the given domain is 180° < θ < 360°. Thus, the cosθ < 0.

$\large{cos \theta = \cancel\frac{ \sqrt{17} }{17} \longrightarrow cos \theta = - \frac{ \sqrt{17} }{17}}$

Then use the Identity of sinθ = tanθcosθ to find the sinθ.

$\large{sin \theta = 4 \times ( - \frac{ \sqrt{17} }{17}) } \\ \large{sin \theta = - \frac{4 \sqrt{17} }{17} }$

Answer

sinθ = -4sqrt(17)/17 or A choice.

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3 years ago